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A line that touches
A STORY
Where the wheel meets the road
Arjun holds his bicycle steady by the handlebar. Tara crouches by the front wheel, pointing low at the tyre.
“Only one point of this wheel is on the road,” Tara says. “Right here, where I am pointing.”
“And the hub sits straight above that point,” Arjun says. “The line from the hub down to it meets the road at a right angle.”
“Every wheel on flat ground does that,” Tara says. “A small wheel or a big one.”
“Now suppose I stand at one spot on the road, away from the wheel,” Arjun says. “Could I draw two straight lines from my spot, each just touching the wheel?”
“Exactly two,” Tara says. “The road itself is one of them. And the two lines turn out equal in length, from your spot to where each one touches.”
Arjun looks from the wheel to the road and back.
Why should those two lengths be equal?
Draw a circle. Then draw a straight line that just brushes past it. Slide the line a little. It now either misses the circle, or cuts through it at two points. At one exact position only, the line touches the circle at a single point and nowhere else. That one line is what this chapter studies.
A line meeting a circle at exactly one point is a TANGENT. The single shared point is the POINT OF CONTACT. A tangent is always perpendicular to the radius drawn to its point of contact. Hold onto that one right angle. We will use it in every proof and every worked example ahead.
Now pick any point outside the circle. From that point, we can always draw exactly two tangents back to the circle. They touch the circle at two different places. Draw both tangents, then measure their lengths with a ruler. The two lengths always come out equal.
Five points of contact spaced unevenly around one circle, each with the same right angle between its radius and its tangent line.
A circle with centre C and a tangent at P, on a coordinate grid, with a small square marking the right angle at P.
Check yourself
A tangent to a circle is best described as a line that
meets the circle at exactly two points
always passes through the centre
never touches the circle at all
meets the circle at exactly one point
Check your answer
meets the circle at exactly two points — Meeting a circle at two points is the definition of a secant, not a tangent.
always passes through the centre — A tangent need not pass through the centre — only the radius to the point of contact does.
never touches the circle at all — A line that never touches the circle meets it at zero points, which is neither a secant nor a tangent.
✓ meets the circle at exactly one point — (D) A tangent is defined by touching the circle at exactly one point, its point of contact.
At the point where a tangent touches a circle, the tangent and the radius drawn to that point are
perpendicular to each other
parallel to each other
equal in length
at $45^\circ$ to each other
Check your answer
✓ perpendicular to each other — (A) The radius to the point of contact always meets the tangent there at a right angle.
parallel to each other — A radius and its tangent meet at the point of contact, so they cannot be parallel lines.
equal in length — The radius-tangent relationship is about the angle between them, not about matching lengths.
at $45^\circ$ to each other — The chapter fixes this angle at exactly $90^\circ$, never at $45^\circ$.
From a point $Q$ outside a circle, two tangents $Q A$ and $Q B$ touch the circle at $A$ and $B$. What can be said about $Q A$ and $Q B$ without any further calculation?
$Q A > Q B$, since one tangent is usually the longer one
$Q A = Q B$ only if $Q$ lies on a special line
nothing can be said until the radius is known
$Q A = Q B$
Check your answer
$Q A > Q B$, since one tangent is usually the longer one — The frame states the two tangents from one external point are equal, not merely close in length.
$Q A = Q B$ only if $Q$ lies on a special line — The equal-length fact holds for every external point, with no extra condition on where $Q$ sits.
nothing can be said until the radius is known — The equal-tangent-length fact follows from the point being external — no radius value is needed.
✓ $Q A = Q B$ — (D) Two tangents drawn from one external point are always equal in length, by the chapter’s own frame.
You cannot always draw a triangle to prove a circle fact. Two straight lines from outside do the same job. Try each check below.
RHS congruence (Class 9, Ganita Manjari, page 104). Here: right triangles $O P T$, $O Q T$ share $O T$ and have equal radii. Check: Hypotenuse $13$ and one leg $5$ leave the other leg $12$.
Perpendicular bisects a chord (Class 9, Ganita Manjari, page 102). Here: point $P$ splits the chord $A B$ into two equal halves this same way. Check: A chord $3$ cm from radius $5$ has half-length $4$: $3^2 + 4^2 = 5^2$.
Opposite sides equal (Class 8, Ganita Prakash Part 1, page 98). Here: the proof starts from $A B = C D$ and $A D = B C$, already true here. Check: A parallelogram with sides $7$ and $9$ has perimeter $2(7 + 9) = 32$.
Isosceles base angles (Class 8, Ganita Prakash Part 1, page 87). Here: triangle $T P Q$ is isosceles, so its base angles are equal. Check: Sides $6, 6$ cm and top angle $40^\circ$ give base angles $70^\circ$ each.
Pythagoras’ theorem (Class 8, Ganita Prakash Part 2, page 44). Here: the tangent length $P Q$ comes from $O Q^2 = O P^2 + P Q^2$. Check: $5^2 + 12^2 = 13^2$, since $25 + 144 = 169$.
Quadrilateral angle sum (Class 8, Ganita Prakash Part 1, page 95). Here: quadrilateral $O P T Q$ uses this to find $\angle P T Q$. Check: Angles $90^\circ, 90^\circ, 130^\circ$ leave a fourth of $50^\circ$, adding to $360^\circ$.
If any of these felt new, read the page named before going on.
Three panels show the same circle with a line missing it, crossing it at two points, and resting on top at one point.
Take any straight line, and any circle drawn in the same plane. Between them, only three things can happen. The line can miss the circle completely. It can cross the circle at two separate points. Or it can touch the circle at exactly one point.
A line meeting the circle at two points is a SECANT. A line meeting it at exactly one point is a TANGENT. That single meeting point is the POINT OF CONTACT.
Try it yourself. Draw a line through two points that sit close together. You will find it is still a secant, however small the gap. It only becomes a tangent once those two points are exactly the same point.
Check yourself
A line meeting a circle at exactly two points is called a
tangent
chord
secant
diameter
Check your answer
tangent — A tangent meets the circle at exactly one point, not two.
chord — A chord is the segment joining the two points; the secant is the whole line through them.
✓ secant — (C) A secant is defined as a line meeting the circle at exactly two points.
diameter — A diameter is a chord through the centre — one special case, not the general secant.
A student says: “A line and a circle can meet at two points, touch at one point, or not meet at all — those are the only three possibilities.” Is this correct?
Yes — those are the only three cases
No — a line can also meet a circle at three points
No — every line must meet the circle at least once
Check your answer
✓ Yes — those are the only three cases — (A) A line and a circle in one plane meet in exactly one of three ways: not at all, at two points, or at one point.
No — a line can also meet a circle at three points — A straight line can meet a circle at most at two points; a third intersection is not possible.
No — every line must meet the circle at least once — A line can also miss the circle completely — that is the third of the three cases, not an impossible one.
A tangent’s single point of contact is given a name distinct from a secant’s two points. Why does it need a separate name?
because tangents are always longer than secants
because a tangent must always be drawn from the centre
it names the one single-point case
because a secant never actually touches the circle
Check your answer
because tangents are always longer than secants — Neither tangents nor secants have a fixed length by definition — length is not what separates the two names.
because a tangent must always be drawn from the centre — A tangent is drawn from outside the circle, touching it at one point; the centre plays no role in drawing it.
✓ it names the one single-point case — (C) The point of contact names the special one-point case, distinct from the two crossing points of a secant.
because a secant never actually touches the circle — A secant does touch the circle, at two points; it is a tangent that touches at only one.
Picture a secant sliding across a circle. Its two crossing points drift toward each other. Watch closely as they move closer. The line still crosses the circle at two separate places. It is still a secant.
A tangent is what a secant becomes only at the exact moment its two points merge into one. Not one instant before. However close the two points remain, while still distinct, the line stays a secant.
This is why a tangent is called the LIMITING POSITION of a secant. The secant approaches the tangent as its two points approach each other. You only get a tangent at that single moment of exact coincidence.
Three secants cross the circle twice each, closer together as the line rises, and a fourth line touches it once, at the tangent.
A SCHOLAR INDIA REMEMBERS
You say a tangent touches the circle at one point. What is your reason? Watch a secant. It cuts the circle at two points. Slide one point towards the other. The two points come closer, and at the end they meet. The line through that one point is the tangent.
Gautama · the sarus crane
Check yourself
As a secant is moved so its two intersection points move closer together, at what stage does the line become a tangent?
only when the two points coincide exactly into one point
as soon as the two points are too close to tell apart
once the gap is smaller than the circle’s radius
the line stays a secant forever and never becomes a tangent
Check your answer
✓ only when the two points coincide exactly into one point — (A) A secant becomes a tangent only at the exact moment its two intersection points merge into a single point.
as soon as the two points are too close to tell apart — However close two points are, the line stays a secant until they coincide exactly.
once the gap is smaller than the circle’s radius — There is no distance threshold in this definition — only exact coincidence turns a secant into a tangent.
the line stays a secant forever and never becomes a tangent — At exact coincidence of the two points, the line does become a tangent — that is the whole point of the limit idea.
Why is “the two points are very close together” not an acceptable description of a tangent?
distinct points, however close, still make a secant
because a tangent must always be drawn outside the circle
because closeness between two points on a circle cannot be measured
because a secant always has fewer points than a tangent
Check your answer
✓ distinct points, however close, still make a secant — (A) Two points that remain distinct, no matter how close, still make the line a secant, never a tangent.
because a tangent must always be drawn outside the circle — Where a tangent is drawn from has nothing to do with why closeness fails as a test.
because closeness between two points on a circle cannot be measured — The distance between two points on a circle can be measured like any other distance; that is not the issue.
because a secant always has fewer points than a tangent — A secant has more points on the circle than a tangent, two against one, not fewer.
A line meets a circle at two points $0.0001$ mm apart. Is this line a tangent?
Yes — the distance is small enough to treat the line as a tangent
It depends on the radius of the circle
No — it is still a secant
Yes, because the line only meaningfully touches the circle once
Check your answer
Yes — the distance is small enough to treat the line as a tangent — A secant does not turn into a tangent by having a small gap; only exact coincidence of the two points does that.
It depends on the radius of the circle — Whether a line is a secant or a tangent depends only on how many points it shares with the circle, never on the radius.
✓ No — it is still a secant — (C) However small the gap, two distinct intersection points still make the line a secant, not a tangent.
Yes, because the line only meaningfully touches the circle once — Geometrically the line still meets the circle at two separate points, so it remains a secant regardless of how the gap looks in practice.
How many tangents can you draw to a circle from a chosen point? The answer depends on where the point sits.
Inside the circle: no tangent exists at all. Every line through an interior point crosses the circle twice, so it is always a secant.
On the circle: exactly one tangent can be drawn, at that point itself.
Outside the circle: exactly two tangents can be drawn, touching the circle at two different points.
Your turn: a point sits exactly on the circle. How many tangents pass through it? (Answer: exactly one, at that point.)
Three panels place a point Q inside, on, and outside the same circle, showing how many tangents can be drawn from each position.
Check yourself
From a point lying INSIDE a circle, how many tangents can be drawn to the circle?
none
exactly one
exactly two
infinitely many
Check your answer
✓ none — (A) No tangent can be drawn from an interior point, since every line through it must cross the circle twice.
exactly one — Exactly one tangent is possible from a point ON the circle, not from a point strictly inside it.
exactly two — Two tangents are possible from a point OUTSIDE the circle; an interior point has none at all.
infinitely many — The chapter fixes the count at zero, one, or two, depending on location — never an infinite number.
From a point lying ON a circle, how many tangents can be drawn at that point?
none
exactly one
exactly two
it depends on the radius
Check your answer
none — Zero tangents is the count from an INTERIOR point; a point on the circle always has one.
✓ exactly one — (B) A point on the circle has exactly one tangent, drawn at that very point.
exactly two — Two tangents come from a point OUTSIDE the circle; a point on the circle itself has only one.
it depends on the radius — The tangent count depends only on where the point sits relative to the circle, never on the radius’s size.
Why can no tangent be drawn from a point strictly inside a circle?
the point sits too close to the centre for a straight line to reach out
tangents can only start from outside the circle, by definition
the radius blocks any line from leaving the circle at that point
every line through it crosses the circle twice
Check your answer
the point sits too close to the centre for a straight line to reach out — A straight line can always be drawn through any interior point; the issue is what kind of line it becomes, not whether it can be drawn.
tangents can only start from outside the circle, by definition — This restates the fact rather than explaining it — the real reason is that every line through an interior point crosses the circle twice.
the radius blocks any line from leaving the circle at that point — A radius does not block a line; the real reason is that an interior-point line always meets the circle twice, making it a secant.
✓ every line through it crosses the circle twice — (D) Any line drawn through an interior point necessarily crosses the circle at two points, making it a secant.
A point $R$ lies outside a circle. How many tangent lines touch the circle from $R$, and how are their lengths related?
exactly two tangents, but their lengths are usually different
exactly one tangent, since only one line can touch the circle from outside
infinitely many tangents, all different lengths
exactly two tangents, and their lengths from $R$ are equal
Check your answer
exactly two tangents, but their lengths are usually different — The two tangent lengths from one external point are always equal, not merely usually equal.
exactly one tangent, since only one line can touch the circle from outside — An external point has exactly two tangents, not one; only a point ON the circle has just one.
infinitely many tangents, all different lengths — An external point has exactly two tangent lines to the circle, never an infinite number.
✓ exactly two tangents, and their lengths from $R$ are equal — (D) An external point always has exactly two tangents, and the two tangent lengths from it are equal.
Because the two lengths always agree, a single number is enough to describe how far a point is from the circle along a tangent.
Take a point $P$ outside a circle, with one tangent drawn from it. You can measure exactly one distance here: from $P$ to the point where that tangent touches the circle. That distance is the LENGTH OF THE TANGENT from $P$.
Two tangents can be drawn from $P$, so the term applies separately to each one. $P$ has two tangent lengths: one for the tangent touching the circle at one point, and one for the tangent touching it at the other.
The two tangent lengths from $P$ always come out equal. So you can safely say “the” tangent length from $P$, without naming which of the two tangents you mean.
Your turn: a tangent from a point $P$ touches a circle at a point $6$ cm away. What is the length of that tangent? (Answer: $6$ cm, the distance from $P$ to the point of contact.)
Check yourself
The “length of the tangent” from an external point $P$ means the distance from $P$ to
the centre of the circle
the nearest point on the circle
the tangent’s point of contact
the far side of the circle
Check your answer
the centre of the circle — The distance from $P$ to the centre $O$ is a different quantity from the tangent length itself.
the nearest point on the circle — The nearest point on the circle need not be the point where the tangent actually touches it.
✓ the tangent’s point of contact — (C) Tangent length is measured from the external point to that tangent’s own point of contact.
the far side of the circle — The far side of the circle plays no part in this definition; only the point of contact does.
From an external point $P$, two tangents touch a circle at $Q$ and $R$. How many different “tangent lengths” does $P$ have, and what are they?
two — $P Q$ and $P R$, one for each tangent from $P$
one — both tangents from the same point must share a single length
three — $P Q$, $P R$, and the distance $P O$ to the centre
none — length only makes sense for a chord, not a tangent
Check your answer
✓ two — $P Q$ and $P R$, one for each tangent from $P$ — (A) Since two tangents are drawn from $P$, the term “tangent length” applies separately to each: $P Q$ and $P R$.
one — both tangents from the same point must share a single length — That the two lengths are equal is a theorem to be proved, not something the definition of tangent length assumes.
three — $P Q$, $P R$, and the distance $P O$ to the centre — The distance $P O$ to the centre is not a tangent length at all; only $P Q$ and $P R$ qualify.
none — length only makes sense for a chord, not a tangent — A tangent segment from $P$ to its point of contact is measurable just like any other segment.
A student finds the tangent length from an external point $P$ by measuring the distance from $P$ to the CENTRE of the circle. What is wrong with this method?
it measures to the centre, not the point of contact
nothing is wrong, since both distances are always equal
the method fails only because $P$ must lie inside the circle
the method fails only when the circle’s radius is unknown
Check your answer
✓ it measures to the centre, not the point of contact — (A) Tangent length is defined from $P$ to the point of contact, a different segment from $P$ to the centre.
nothing is wrong, since both distances are always equal — The distance to the centre and the tangent length are different segments and are not equal in general.
the method fails only because $P$ must lie inside the circle — $P$ must lie outside the circle for a tangent to exist at all; that is not the error in this method.
the method fails only when the circle’s radius is unknown — The method is wrong regardless of whether the radius is known — it measures the wrong segment either way.
Two right triangles share a hypotenuse of 13 and a leg of 5, forcing the third side to be 12 in both.
RHS congruence is a shortcut for right triangles only. The name gives the checklist: Right angle, Hypotenuse, Side.
Let us check the right angle first. Both triangles must carry one. Next, we match the HYPOTENUSE, the side opposite the right angle, between the two triangles. Then we match one more side, any one we like.
SSS, SAS, ASA and RHS between them cover every pair of congruent triangles. Only RHS is used here, to show that the two tangent lengths from an outside point are equal.
Check yourself
The RHS congruence criterion proves two triangles congruent when
an equal hypotenuse plus one equal side, in right triangles
all three corresponding sides equal, with no angle needed at all
two sides and the angle between them equal, with no right angle needed
two angles and the side between them equal, with no right angle needed
Check your answer
✓ an equal hypotenuse plus one equal side, in right triangles — (A) RHS needs both triangles to be right triangles, with matching hypotenuses and one further matching side.
all three corresponding sides equal, with no angle needed at all — Matching all three sides is the SSS criterion, a different route to congruence that needs no right angle at all.
two sides and the angle between them equal, with no right angle needed — Two sides and the angle between them is the SAS criterion, which works on any triangle, not only right ones.
two angles and the side between them equal, with no right angle needed — Two angles and the side between them is the ASA criterion, which also works on any triangle, not only right ones.
Why does RHS need the two triangles to be right triangles, unlike SSS?
it does not really need a right angle; the name is just a label
right triangles are always larger than other triangles
only right triangles have a hypotenuse worth measuring
the right angle fixes the third side from the other two
Check your answer
it does not really need a right angle; the name is just a label — The right angle is essential to RHS — without it, matching a hypotenuse and one side would not fix the triangle.
right triangles are always larger than other triangles — The size of a triangle has nothing to do with why RHS requires a right angle.
only right triangles have a hypotenuse worth measuring — Every triangle has a longest side; a hypotenuse is simply the name for that side in a right triangle, which is not the reason RHS needs one.
✓ the right angle fixes the third side from the other two — (D) Because of the right angle, the hypotenuse and one side together already fix the length of the third side.
A student applies RHS to two triangles with an equal hypotenuse and one equal side, but neither triangle has a right angle. Is this application valid?
No — RHS requires both triangles to be right triangles first
Yes — a matching hypotenuse and one side is enough regardless of angles
Yes, provided the two triangles have the same perimeter
No — RHS additionally requires all three sides to match
Check your answer
✓ No — RHS requires both triangles to be right triangles first — (A) RHS only applies once both triangles are already known to be right triangles.
Yes — a matching hypotenuse and one side is enough regardless of angles — Without a right angle in each triangle, there is no hypotenuse to speak of, and RHS cannot apply.
Yes, provided the two triangles have the same perimeter — Matching perimeters plays no role in RHS; a right angle in each triangle is what the criterion actually needs.
No — RHS additionally requires all three sides to match — RHS needs only the hypotenuse and one further side to match, not all three sides as SSS does.
A circle with centre O and radius r to the point P, where a tangent line rests, with a square marking the right angle there.
THEOREM. The tangent to a circle, at the point of contact, is perpendicular to the radius drawn to that point.
Let $O$ be the centre of a circle. Let $X Y$ be a tangent to the circle, touching it at the point $P$. We must show that $O P$ is perpendicular to $X Y$.
Before the proof, let us see what it compares. It compares the length of $O P$ against the distance from $O$ to every other point on $X Y$. It shows $O P$ is the SHORTEST such distance, and only a perpendicular can be the shortest one. Follow the numbered proof below, one justification per step, and check each line against the figure as we go.
The proof never measures an angle: it shows OP is the shortest segment, and the shortest is the perpendicular.
Proof
Given. A circle has centre $O$. Its tangent $X Y$ touches the circle at the point $P$. $O P$ is the radius drawn to that point of contact.
To prove. The tangent to a circle at any point is perpendicular to the radius through the point of contact: $O P$ is perpendicular to $X Y$.
let $O$ be the centre of the circle and $X Y$ a tangent touching it at the point $P$; draw the radius $O P$ This is the construction the whole proof turns on.
let $Q$ be any point on the tangent $X Y$ other than $P$, and join $O Q$ We draw this auxiliary line so we can compare the two distances.
$Q$ lies outside the circle A tangent meets the circle at exactly one point, $P$. So any other point of $X Y$ cannot lie on the circle.
$O Q$ is longer than $O P$ $O P$ reaches a point ON the circle, while $O Q$ reaches a point OUTSIDE it, by the step before.
$O Q$ is longer than $O P$ for every choice of $Q$ on $X Y$ other than $P$ Step 4 did not depend on which point $Q$ we chose.
the shortest segment from a point to a line is the perpendicular from that point to the line We already know this fact from Class 9.
$O P$ is perpendicular to $X Y$ By step 5, $O P$ is shorter than the segment to every other point of $X Y$, so it is the shortest such segment. By step 6, the shortest segment is the perpendicular.
■
A SCHOLAR INDIA REMEMBERS
Gautama’s way of arguing, from around the 2nd century BCE, was used by schools that agreed on almost nothing else. They disagreed about many big questions. Still, they could all check the same steps of an argument. A proof in geometry works like that. You may doubt the result at first. You can still check each step, one at a time, and agree when every step holds.
Gautama · the sarus crane
Check yourself
In the proof that a tangent is perpendicular to the radius, why must $Q$, any other point on the tangent besides $P$, lie outside the circle?
$Q$ was deliberately chosen to sit far away from the point $P$
the radius $O P$ happens to be shorter than the circle’s diameter
the tangent meets the circle only at $P$
tangents to a circle are always constructed to lie outside it
Check your answer
$Q$ was deliberately chosen to sit far away from the point $P$ — $Q$ can be chosen anywhere else on the tangent; it lies outside the circle by definition of a tangent, not by how far it is chosen from $P$.
the radius $O P$ happens to be shorter than the circle’s diameter — The radius being shorter than the diameter is true of every circle and has nothing to do with why $Q$ is outside it.
✓ the tangent meets the circle only at $P$ — (C) A tangent touches the circle only at $P$, so any other point on it, including $Q$, cannot also lie on the circle.
tangents to a circle are always constructed to lie outside it — The tangent line touches the circle at $P$ and otherwise runs outside it; that is exactly the fact this step proves, not one it assumes.
Suppose the line $X Y$ through $P$ were a secant instead of a tangent. Which step of the perpendicularity proof breaks down first?
the step joining $O Q$, since $O Q$ cannot be drawn for a secant
the final step invoking the shortest-segment fact, since it only applies to tangents
the step claiming every other point $Q$ on the line lies outside the circle
no step breaks down — the same proof works unchanged for a secant
Check your answer
the step joining $O Q$, since $O Q$ cannot be drawn for a secant — The segment $O Q$ can always be drawn to any point $Q$ on any line; that is not where a secant breaks the proof.
the final step invoking the shortest-segment fact, since it only applies to tangents — The shortest-segment fact is a general geometric fact about any point and line; the proof fails earlier, at the claim that $Q$ lies outside the circle.
✓ the step claiming every other point $Q$ on the line lies outside the circle — (C) A secant has a second point on the circle itself, so the claim that every other point on the line lies outside the circle fails immediately.
no step breaks down — the same proof works unchanged for a secant — The whole proof rests on every other point of the line lying outside the circle, which is false for a secant.
The proof relies on the fact that the shortest segment from a point to a line is the perpendicular from that point. What role does this fact play?
it turns the comparison into a right angle
it proves that $O$ lies on the tangent line
it proves that the tangent has a fixed length
it is only needed to define what a radius is
Check your answer
✓ it turns the comparison into a right angle — (A) This fact converts the earlier distance comparison into the perpendicularity conclusion the proof is aiming for.
it proves that $O$ lies on the tangent line — This fact says nothing about $O$ lying on the line; $O$ is the centre, off the tangent entirely.
it proves that the tangent has a fixed length — The shortest-segment fact is about direction (perpendicularity), not about fixing any particular length.
it is only needed to define what a radius is — The radius is already defined earlier in the proof; this fact’s real job is converting a distance comparison into a right angle.
The theorem proves the radius is perpendicular to the tangent at the point of contact. Which of these does this result rule out?
a circle having more than one tangent at a single point
any tangent-radius angle other than $90^\circ$
a chord being perpendicular to a diameter
two tangents from an external point being unequal in length
Check your answer
a circle having more than one tangent at a single point — How many tangents a circle has at one point is a separate fact from the angle a tangent makes with the radius.
✓ any tangent-radius angle other than $90^\circ$ — (B) The theorem forces the radius-tangent angle at the point of contact to be exactly $90^\circ$, ruling out every other angle there.
a chord being perpendicular to a diameter — A chord’s relationship to a diameter is a different, unrelated fact from the tangent-radius angle this theorem fixes.
two tangents from an external point being unequal in length — Equal tangent lengths from an external point is a separate theorem, proved later using this one, not the same claim.
A line $l$ touches a circle at $P$, and the radius $O P$ makes an angle of $80^\circ$ with $l$. What can you conclude?
$l$ is not a tangent to the circle at $P$
$l$ is a tangent, since $80^\circ$ is close enough to a right angle
$l$ is a tangent only if the circle’s radius is large enough
nothing can be concluded without knowing the circle’s centre
Check your answer
✓ $l$ is not a tangent to the circle at $P$ — (A) Since a genuine tangent must meet its radius at exactly $90^\circ$, an $80^\circ$ angle means $l$ is not actually a tangent there.
$l$ is a tangent, since $80^\circ$ is close enough to a right angle — The theorem requires exactly $90^\circ$, not an angle merely close to it, so $l$ cannot be a tangent here.
$l$ is a tangent only if the circle’s radius is large enough — Whether $l$ is a tangent depends only on the angle at the point of contact, never on the size of the radius.
nothing can be concluded without knowing the circle’s centre — The theorem alone is enough: any angle other than $90^\circ$ at the point of contact rules out a genuine tangent.
Radii OQ and OR carry matching ticks, tangents from P touch Q and R sharing segment OP, with right angles at Q and R.
THEOREM. The lengths of the two tangents drawn from an external point to a circle are equal.
Let $O$ be the centre of a circle. Let $P$ be a point outside it, from which two tangents touch the circle at $Q$ and $R$. We must show that $P Q$ equals $P R$.
Let us watch for the shape the proof builds: two right triangles sharing one common hypotenuse. We compare them using the RHS congruence criterion we just met. The equal tangent lengths then follow as corresponding sides of congruent triangles. Read the numbered proof below, one justification per step, and see the same shape appear there.
The two right triangles OQP and ORP share a radius, a right angle and the hypotenuse OP, so PQ equals PR.
Proof
Given. A point $P$ sits outside a circle with centre $O$. Two tangents $P Q$ and $P R$ touch the circle at $Q$ and $R$. The radii $O Q$ and $O R$ go to the two points of contact, and $O P$ is joined.
To prove. The two tangents drawn to a circle from an external point are equal in length: $P Q = P R$.
let $P$ be a point outside the circle, with tangents $P Q$ and $P R$ touching it at $Q$ and $R$. Draw the radii $O Q$ and $O R$, and join $O P$ This is the construction the proof compares.
$O Q = O R$ Both are radii of the same circle.
$\angle O Q P = \angle O R P = 90^\circ$ Each is the angle between a radius and the tangent at its own point of contact. The theorem before this one makes every such angle a right angle.
$O P = O P$ This is the same segment, common to both triangles.
$\triangle O Q P \sim \cdot \equiv \triangle O R P$ By the RHS criterion: the hypotenuse $O P$ is common, and the radii $O Q = O R$ are equal.
$P Q = P R$ Corresponding sides of congruent triangles are equal. This is CPCT.
■
Check yourself
In the proof that $P Q = P R$, why are $O Q$ and $O R$ equal?
because both are tangents from the same point
because both are radii of the same circle
because the two triangles are given as congruent from the start
because $Q$ and $R$ are the same distance from $P$
Check your answer
because both are tangents from the same point — $O Q$ and $O R$ are radii, drawn from the centre to the points of contact — the tangents are $P Q$ and $P R$ instead.
✓ because both are radii of the same circle — (B) $O Q$ and $O R$ are both radii of the same circle, so they are equal by definition.
because the two triangles are given as congruent from the start — The congruence of the two triangles is what the proof is trying to establish, not something assumed at this early step.
because $Q$ and $R$ are the same distance from $P$ — That $P Q$ equals $P R$ is the theorem’s conclusion, not a fact available this early in the proof.
Which earlier theorem supplies the two right angles the RHS step needs?
the theorem that opposite angles of a cyclic quadrilateral are supplementary
the fact that all radii of a circle are equal
the tangent-perpendicular-to-radius theorem
the definition of a secant
Check your answer
the theorem that opposite angles of a cyclic quadrilateral are supplementary — Cyclic quadrilaterals are not part of this proof; the right angles here come from the tangent-radius theorem instead.
the fact that all radii of a circle are equal — Equal radii give $O Q = O R$, a different step; the right angles come from the tangent-perpendicular-to-radius theorem.
✓ the tangent-perpendicular-to-radius theorem — (C) The right angles at $Q$ and $R$ come directly from the earlier tangent-perpendicular-to-radius theorem.
the definition of a secant — A secant’s definition plays no role in fixing a right angle; that comes from the earlier tangent theorem.
Suppose $P Q$ and $P R$ were secants through $P$ rather than tangents. Which step of the equal-tangent-lengths proof fails first?
the step claiming $\angle O Q P = \angle O R P = 90^\circ$
the step $O P = O P$, since a common side cannot be assumed for secants
the step $O Q = O R$, since radii are not equal for secants
no step fails — RHS applies to any two lines through $P$
Check your answer
✓ the step claiming $\angle O Q P = \angle O R P = 90^\circ$ — (A) That right angle comes only from the tangent-radius theorem, which no longer applies once the lines are secants.
the step $O P = O P$, since a common side cannot be assumed for secants — $O P$ equals itself regardless of what kind of lines pass through $P$; that step never depends on tangency.
the step $O Q = O R$, since radii are not equal for secants — $O Q$ and $O R$ are still radii of the same circle and remain equal no matter what kind of line passes through $Q$ or $R$.
no step fails — RHS applies to any two lines through $P$ — Without the right angle a tangent guarantees, RHS has no hypotenuse-and-right-angle setup left to work with.
The equal-tangent-lengths theorem is proved using RHS congruence. Which of these does this result rule out?
two different external points having tangents of different lengths
a tangent and a secant from the same point having equal length
two tangents from the same external point having different lengths
the two tangents making unequal angles with the line to the centre
Check your answer
two different external points having tangents of different lengths — The theorem compares the two tangents from ONE external point; it says nothing about tangents from two different points.
a tangent and a secant from the same point having equal length — The theorem is about two tangents from one point, not about comparing a tangent to a secant.
✓ two tangents from the same external point having different lengths — (C) The theorem forces the two tangent lengths from one external point to be equal, ruling out any difference between them.
the two tangents making unequal angles with the line to the centre — The angle each tangent makes with the line to the centre is a separate result, proved afterward using this same congruence.
Two tangents from an external point $P$ touch a circle at $Q$ and $R$, with $P Q = 8$ cm. What is $P R$?
$8$ cm, but only if $P$ lies on a special line through the centre
$8$ cm, since tangents from one point are equal
it cannot be found without knowing the radius
less than $8$ cm, since $R$ is a different point from $Q$
Check your answer
$8$ cm, but only if $P$ lies on a special line through the centre — The equal-tangent-lengths theorem holds for every external point, with no extra condition on where $P$ sits.
✓ $8$ cm, since tangents from one point are equal — (B) By the equal-tangent-lengths theorem, $P R$ must equal $P Q$, which is $8$ cm.
it cannot be found without knowing the radius — The theorem alone gives $P R = P Q = 8$ cm; the radius is not needed to compare the two tangent lengths.
less than $8$ cm, since $R$ is a different point from $Q$ — $Q$ and $R$ being different points of contact does not make the two tangent lengths different — the theorem says they are equal.
A SCHOLAR INDIA REMEMBERS
Walk me through the steps again, slowly. Join the centre to the outside point. Join the centre to each point of contact. Each radius meets its tangent at $90^\circ$. The two right triangles share their hypotenuse, and their radii are equal. So the triangles are congruent, and the two tangents are equal.
For two tangents $P Q$ and $P R$ drawn from an external point $P$, the line $O P$ to the centre
is perpendicular to $Q R$ but does not bisect the angle
bisects the tangent lengths $P Q$ and $P R$ instead
has no fixed relationship to $\angle Q P R$
bisects $\angle Q P R$, the angle between the tangents
Check your answer
is perpendicular to $Q R$ but does not bisect the angle — $O P$ does bisect $\angle Q P R$ — that is exactly what this fact states, following from the same congruence used for the tangent lengths.
bisects the tangent lengths $P Q$ and $P R$ instead — $P Q$ and $P R$ are already equal in full, not split in half by $O P$; it is the ANGLE between them that $O P$ bisects.
has no fixed relationship to $\angle Q P R$ — $O P$ has a fixed relationship to $\angle Q P R$ — it bisects it, by the same congruence that proves the tangents equal.
✓ bisects $\angle Q P R$, the angle between the tangents — (D) The line from the external point to the centre always bisects the angle between the two tangents.
Why does the same congruence, $\triangle O Q P \sim \cdot \equiv \triangle O R P$, give both equal tangent lengths AND equal angles at $P$?
tangent lengths and angles are always numerically equal to each other
it is a coincidence specific to this one theorem
$O P$ is drawn twice, once for each fact
congruent triangles match in every corresponding part
Check your answer
tangent lengths and angles are always numerically equal to each other — A length and an angle are different kinds of quantity and are never claimed to be numerically equal here.
it is a coincidence specific to this one theorem — This is not a coincidence — it is exactly what corresponding-parts-of-congruent-triangles guarantees, every time.
$O P$ is drawn twice, once for each fact — $O P$ is drawn only once; both facts come from the SAME single congruence, not from drawing it again.
✓ congruent triangles match in every corresponding part — (D) Congruent triangles match in every corresponding part, so both the sides and the angles come out equal together.
In a figure, $\angle O P Q = 35^\circ$. What is $\angle Q P R$, the angle between the two tangents?
$35^\circ$, the same as $\angle O P Q$
$110^\circ$, from a supplementary relationship
it cannot be found without knowing the radius
$70^\circ$, since $O P$ bisects $\angle Q P R$
Check your answer
$35^\circ$, the same as $\angle O P Q$ — $\angle O P Q$ is only HALF of $\angle Q P R$, since $O P$ bisects it; the full angle is double, not equal.
$110^\circ$, from a supplementary relationship — There is no supplementary relationship in play here — doubling the half-angle is the correct step, giving $70^\circ$.
it cannot be found without knowing the radius — The bisector fact alone is enough: doubling $\angle O P Q$ gives $\angle Q P R$, with no radius needed.
✓ $70^\circ$, since $O P$ bisects $\angle Q P R$ — (D) Since $O P$ bisects the angle between the tangents, $\angle Q P R$ is twice $\angle O P Q$, giving $70^\circ$.
The same circle and tangents, now with equal tangent lengths marked, and two equal arcs at P splitting the angle into two equal halves.
Let us look back at the congruence we just used to prove the two tangent lengths equal. It gives us one more fact for free.
Triangle $O Q P$ is congruent to triangle $O R P$. So every corresponding pair of angles is equal too, not only the sides. In particular, $\angle O P Q = \angle O P R$.
*So the line $O P$, from the centre to the external point, bisects the angle between the two tangents.* We get this from CPCT, applied a second time to the same congruence. No new construction is needed.
Your turn: if $\angle O P Q = 35^\circ$, what is $\angle O P R$? (Answer: $35^\circ$, since $O P$ bisects the angle between the tangents.)
A student says: “A tangent touches a circle at two points that are extremely close together.” Is this correct?
Yes — when two points are close enough, the line counts as a tangent
Yes, but only for very small circles
No — however close two points are, the line is still a secant
No — a tangent actually touches the circle at zero points
Check your answer
Yes — when two points are close enough, the line counts as a tangent — Only exact coincidence of the two points makes a tangent; being close is never sufficient, however small the gap.
Yes, but only for very small circles — The definition of a tangent does not depend on the size of the circle at all; closeness never turns a secant into a tangent, at any size.
✓ No — however close two points are, the line is still a secant — (C) A tangent meets the circle at exactly one point; two points, however close, still make the line a secant.
No — a tangent actually touches the circle at zero points — A tangent does touch the circle, at exactly one point; zero points would describe a line that misses the circle entirely.
What exactly has to happen to a secant’s two intersection points for the line to become a tangent?
the two points must move to within a hundredth of the radius
the two points must coincide exactly into a single point
the two points must both move to the centre of the circle
the line must be rotated by exactly $90^\circ$
Check your answer
the two points must move to within a hundredth of the radius — No distance threshold appears anywhere in this definition; only exact coincidence of the two points matters.
✓ the two points must coincide exactly into a single point — (B) Only the exact coincidence of the two intersection points turns a secant into a tangent.
the two points must both move to the centre of the circle — The two points must coincide with EACH OTHER on the circle, not move toward the centre, which is not even on the circle.
the line must be rotated by exactly $90^\circ$ — Rotating the line plays no role here; what matters is the two intersection points meeting exactly, regardless of the line’s angle.
Why is “the two points are indistinguishable to the eye” a poor test for whether a line is a tangent?
because the eye cannot see circles accurately at all
because tangents can only be confirmed using algebra, never a diagram
because a secant and a tangent always look completely different
looking the same is not being exactly the same
Check your answer
because the eye cannot see circles accurately at all — This is not about how well circles can be seen in general; it is about the gap between looking the same and being exactly the same.
because tangents can only be confirmed using algebra, never a diagram — Diagrams are used throughout this chapter; the actual issue is that visual closeness is not the same as exact coincidence.
because a secant and a tangent always look completely different — A secant with two very close points can look just like a tangent, which is exactly why a visual test is unreliable.
✓ looking the same is not being exactly the same — (D) Two points can look identical to the eye while remaining mathematically distinct, so this is not a valid test for tangency.
A line meets a circle at two points $1$ mm apart on a circle of radius $10$ cm. A second line meets the same circle at exactly one point. Which line, if either, is the tangent?
only the second line — the first still has two distinct points
both lines are tangents, since $1$ mm is negligible next to $10$ cm
only the first line, since it comes closer to touching at one point in practice
neither line is a tangent unless the points coincide with the centre
Check your answer
✓ only the second line — the first still has two distinct points — (A) The first line still has two distinct intersection points, however small the gap, so only the second line, with one point, is the tangent.
both lines are tangents, since $1$ mm is negligible next to $10$ cm — The size of the gap relative to the radius is irrelevant; the first line still has two distinct points, so it remains a secant.
only the first line, since it comes closer to touching at one point in practice — The SECOND line is the tangent, since it meets the circle at exactly one point; the first line still has two, however close.
neither line is a tangent unless the points coincide with the centre — Coinciding with the centre plays no role in this definition; a tangent only needs its intersection points to coincide with EACH OTHER.
A student argues that as a secant’s two points get closer, the line “gradually becomes more of a tangent,” so tangency is a matter of degree. What is wrong with this claim?
nothing is wrong; a line can be partly a tangent and partly a secant
tangency is not a matter of degree
the claim is wrong only because secants cannot have points close together
the claim is right, but only for circles larger than a certain size
Check your answer
nothing is wrong; a line can be partly a tangent and partly a secant — A line is fully one or fully the other, secant or tangent — there is no partly-tangent category in this definition.
✓ tangency is not a matter of degree — (B) A line stays fully a secant right up until its two points coincide exactly, at which point it becomes fully a tangent, with no in-between stage.
the claim is wrong only because secants cannot have points close together — A secant can absolutely have two points very close together; the real issue is that closeness is still not exact coincidence.
the claim is right, but only for circles larger than a certain size — The claim is wrong regardless of the circle’s size; tangency depends only on exact coincidence of the two points, never on scale.
Two panels show the same arc, with two close but separate points on the left and one coincided point on the right.
You may already be carrying this mix-up.
You might picture a tangent as touching a circle at two points placed very close together, rather than at exactly one point. That picture is wrong, however close the two points seem.
Only when the two points coincide EXACTLY, becoming one single point, does the line become a tangent.
Check your own picture against that rule. The distinction is not about closeness. It is about whether there are two points, or one.
Is a line through two points that sit very close together on a circle a tangent?
Weaker. A line passes through $P$ and a second point $Q$ that sits very near $P$ on the circle. It looks like it touches at one place, so it gets called a tangent. But count the points where the line meets the circle: $P$ and $Q$ are two separate points, however close together, so the line is a secant, not a tangent.
Stronger. The tangent at $P$ meets the circle at $P$ and at no other point at all, however closely you look. One point of contact, not two close ones, is what makes a line a tangent.
A SCHOLAR INDIA REMEMBERS
Two very close points are still two points. So a line through them still cuts the circle twice. What is the reason a tangent is different? A tangent meets the circle at exactly one point. Even when the two points are very close, the line is a secant. It becomes a tangent only when the two points meet.
A right triangle sits at the point of contact. It often gets remembered only as “use Pythagoras”, so the two given lengths get squared and added, no matter which two they are. That is not always the right move.
The right angle sits at the point of contact. The side facing it, from the centre to the outside point, is the hypotenuse. The hypotenuse is the square you subtract from, never a square you add.
Your turn: a radius is $7$ cm, and the distance from the centre to an outside point is $25$ cm. What is the tangent length? (Answer: $24$ cm, since $25^2 - 7^2 = 625 - 49 = 576$.)
First try
Radius $12$ cm, distance to the centre $20$ cm. I squared both and added: $12^2 + 20^2 = 544$, so the tangent is $\sqrt{544} \approx 23.3$ cm.
Second look
The right angle sits at the point of contact, so the distance to the centre is the hypotenuse, the longest side. I should subtract: $20^2 - 12^2 = 256$, so the tangent is $\sqrt{256} = 16$ cm.
The line to the centre is the hypotenuse. Take the square of the radius from its square, never add.
Find the tangent length when the radius is 12 centimetres and the distance to the centre is 20 centimetres.
Weaker. Square and add: $P Q^2 = 20^2 + 12^2 = 544$, so $P Q \approx 23.3$ cm. But $23.3$ cm is longer than the $20$ cm from the outside point to the centre, and a leg of a right triangle is never longer than its hypotenuse.
Stronger. The right angle sits at $P$, so $O Q$ is the hypotenuse: $P Q^2 = O Q^2 - O P^2 = 20^2 - 12^2 = 256$, so $P Q = 16$ cm.
In finding a tangent length from the radius and the distance to the centre, which theorem makes the triangle right-angled in the first place?
the Pythagorean theorem itself supplies the right angle
the tangent-perpendicular-to-radius theorem
the equal-tangent-lengths theorem
no theorem is needed — the triangle is automatically right-angled
Check your answer
the Pythagorean theorem itself supplies the right angle — Pythagoras’s theorem needs a right angle to already exist; it is the tangent-perpendicular-to-radius theorem that supplies it.
✓ the tangent-perpendicular-to-radius theorem — (B) The tangent-perpendicular-to-radius theorem fixes the right angle at the point of contact, setting up the right triangle.
the equal-tangent-lengths theorem — The equal-tangent-lengths theorem compares two tangent lengths; it is a different theorem that fixes the right angle here.
no theorem is needed — the triangle is automatically right-angled — The right angle needs justification, supplied by the tangent-perpendicular-to-radius theorem, not assumed on its own.
A tangent touches a circle of radius $9$ cm at $P$. From an external point $Q$, $O Q = 15$ cm. What is the tangent length $P Q$?
$24$ cm, from adding the two given lengths
$6$ cm, from subtracting the two given lengths
$18$ cm, from doubling the radius
$12$ cm, from $15^2 = 9^2 + P Q^2$
Check your answer
$24$ cm, from adding the two given lengths — Adding $15$ and $9$ ignores the right-angle relationship; Pythagoras gives $P Q = 12$ cm instead.
$6$ cm, from subtracting the two given lengths — Subtracting $9$ from $15$ ignores the right-angle relationship; the correct method squares and subtracts, giving $12$ cm.
$18$ cm, from doubling the radius — Doubling the radius has no basis in this right triangle; Pythagoras on $O Q$, $O P$, and $P Q$ gives $12$ cm.
✓ $12$ cm, from $15^2 = 9^2 + P Q^2$ — (D) Applying the right triangle at $P$: $15^2 = 9^2 + P Q^2$ gives $P Q^2 = 144$, so $P Q = 12$ cm.
A tangent touches a circle of radius $7$ cm at $P$. If the tangent length $P Q$ is $24$ cm, what is the distance $O Q$ from the external point to the centre?
$25$ cm, from $O Q^2 = 7^2 + 24^2$
$31$ cm, from adding the two given lengths
$17$ cm, from subtracting the two given lengths
$24$ cm, the same as the tangent length
Check your answer
✓ $25$ cm, from $O Q^2 = 7^2 + 24^2$ — (A) Applying Pythagoras to the right triangle at $P$: $O Q^2 = 7^2 + 24^2 = 625$, so $O Q = 25$ cm.
$31$ cm, from adding the two given lengths — Adding $7$ and $24$ ignores the right-angle relationship; Pythagoras gives $O Q = 25$ cm instead.
$17$ cm, from subtracting the two given lengths — Subtracting $7$ from $24$ ignores the right-angle relationship; the correct method squares and adds, giving $25$ cm.
$24$ cm, the same as the tangent length — $O Q$ is the hypotenuse of the right triangle, always longer than either leg, including the tangent length $P Q$.
A student sets up $O Q^2 = O P^2 + P Q^2$ for a tangent problem but uses the DIAMETER in place of the radius $O P$. What goes wrong?
it substitutes the wrong length for the radius $O P$
nothing goes wrong, since the diameter and radius give the same answer
the triangle is no longer right-angled
the formula itself changes when the diameter is used
Check your answer
✓ it substitutes the wrong length for the radius $O P$ — (A) $O P$ in this formula is the radius to the point of contact; using the diameter puts the wrong length into the equation.
nothing goes wrong, since the diameter and radius give the same answer — The diameter is twice the radius, so swapping it in changes the equation and gives a wrong answer, not the same one.
the triangle is no longer right-angled — The right angle at $P$ comes from the tangent-radius theorem and is unaffected by which length is substituted; only the ARITHMETIC goes wrong.
the formula itself changes when the diameter is used — The formula $O Q^2 = O P^2 + P Q^2$ stays the same; the error is putting the wrong number, the diameter, into the $O P$ slot.
Worked example
A tangent length from the radius and the distance to the centre
a tangent touches a circle of radius $5$ cm at the point $P$; the distance from the centre $O$ to an external point $Q$ is $O Q = 13$ cm These are the given measurements.
$\angle O P Q = 90^\circ$ The radius to the point of contact is always perpendicular to the tangent there.
$O Q^2 = O P^2 + P Q^2$ Triangle $O P Q$ is right-angled at $P$, so Pythagoras’ theorem applies.
$13^2 = 5^2 + P Q^2$ Substitute the given lengths.
$P Q^2 = 169 - 25 = 144$ Evaluate the squares, then subtract.
$P Q = 12$ cm Take the square root of $144$.
$5^2 + 12^2 = 25 + 144 = 169 = 13^2$ Check the answer by substituting back. $5$, $12$ and $13$ satisfy Pythagoras exactly, so $P Q = 12$ cm is correct.
A right triangle with radius 5 at OP, tangent 12 at PQ, and OQ equal to 13, with a right angle at P.
the tangent
the point of contact
the external point
The taut rope runs from the stake and meets the round tank at one point. It lies along a tangent.
Finding a tangent length from the radius and the distance to the centre
Mark the right angle The radius meets the tangent at $90^\circ$, at the point of contact $P$. Here the radius is $O P = 5$ cm and $\angle O P Q = 90^\circ$.
Name the hypotenuse The side from the centre to the outside point is the hypotenuse. Here that is $O Q = 13$ cm.
Subtract the squares $P Q^2 = O Q^2 - O P^2 = 13^2 - 5^2 = 144$, so $P Q = 12$ cm.
Check the tangent is shorter The tangent must come out shorter than $O Q$. $12$ cm is shorter than $13$ cm, so the answer checks out.
In the kite OPTQ, the right angles at P and Q leave the angle at O, 130 degrees, and at T, 50 degrees.
Worked example
The angle between two tangents, given the angle at the centre
two tangents $T P$ and $T Q$ are drawn to a circle with centre $O$ from an external point $T$, touching it at $P$ and $Q$; $\angle P O Q = 130^\circ$ These are the given measurements.
$\angle O P T = \angle O Q T = 90^\circ$ Each radius meets its own tangent at a right angle.
$\angle P O Q + \angle O P T + \angle P T Q + \angle T Q O = 360^\circ$ The four angles of the quadrilateral $O P T Q$ add to $360^\circ$.
$130^\circ + 90^\circ + 90^\circ + \angle P T Q = 360^\circ$ Substitute $\angle P O Q = 130^\circ$ and the two right angles from step 2.
$\angle P T Q = 50^\circ$ Solve for $\angle P T Q$.
$\angle O T P = 25^\circ$ The line $O T$ bisects the angle between the two tangents, so it splits $\angle P T Q$ into two equal halves.
$130 + 50 = 180$ Check it without redoing the work. The two right angles use up $180^\circ$ of the quadrilateral, so the angle at the centre and the angle between the tangents must always add to $180^\circ$.
Finding the angle between two tangents from the angle at the centre
Mark the two right angles At each point of contact, the radius meets the tangent at $90^\circ$. Here $\angle O P T = \angle O Q T = 90^\circ$.
Add the quadrilateral’s angles The angle at the centre, the two right angles, and the angle at the outside point add to $360^\circ$. Here $130^\circ + 90^\circ + 90^\circ + \angle P T Q = 360^\circ$.
Solve for the angle at T $\angle P T Q = 360^\circ - 130^\circ - 90^\circ - 90^\circ = 50^\circ$.
Halve it for the bisected angle The line from the centre to the outside point bisects that angle, so $\angle O T P = 50^\circ/2 = 25^\circ$.
In proving that $P$ bisects the chord $A B$, why is $O P$ drawn?
$O P$ is the diameter of the smaller circle
$O P$ is equal in length to $A B$
$O P$ is needed only to measure the distance between the two circles
$O P$ is the radius here, and a centre-to-chord line there
Check your answer
$O P$ is the diameter of the smaller circle — $O P$ is a radius of the smaller circle, reaching from the centre to the point of contact, not its diameter.
$O P$ is equal in length to $A B$ — Nothing in the proof claims $O P$ equals the chord $A B$ in length; its role is about direction, not length.
$O P$ is needed only to measure the distance between the two circles — $O P$’s role is to supply a right angle to the chord, using the tangent-radius theorem, not to measure a gap between circles.
✓ $O P$ is the radius here, and a centre-to-chord line there — (D) $O P$ plays a double role — the radius that is perpendicular to the tangent, and a centre-to-chord line for the larger circle.
Which two facts, one about tangents and one about chords, combine to prove $A P = B P$?
the tangent-radius fact, and the Class 9 chord-bisection fact
the equal-tangent-lengths theorem, and the fact that all chords of a circle are equal
the angle-bisector fact, and the definition of a secant
the RHS congruence criterion alone, with no chord fact needed
Check your answer
✓ the tangent-radius fact, and the Class 9 chord-bisection fact — (A) The tangent-radius right angle, combined with the Class 9 chord-bisection fact, together prove $A P = B P$.
the equal-tangent-lengths theorem, and the fact that all chords of a circle are equal — Equal tangent lengths need TWO tangents from one point, which this proof does not use; and chords of a circle are not all equal in general.
the angle-bisector fact, and the definition of a secant — Neither the angle-bisector fact nor the definition of a secant plays any role in proving that $P$ bisects $A B$.
the RHS congruence criterion alone, with no chord fact needed — RHS congruence is not used in this particular proof at all; it needs the tangent-radius fact plus the Class 9 chord-bisection fact.
Two concentric circles share centre $O$. A chord $C D$ of the larger circle touches the smaller circle at $M$, with $C M = 6$ cm. What is $C D$?
$6$ cm, the same as $C M$
$12$ cm, since $M$ bisects $C D$
$18$ cm, from an arbitrary multiple of $C M$
it cannot be found without knowing both radii
Check your answer
$6$ cm, the same as $C M$ — $C M$ is only HALF of the chord $C D$; the whole chord is double that, $12$ cm.
✓ $12$ cm, since $M$ bisects $C D$ — (B) Since $M$ bisects the chord $C D$, $D M$ also equals $6$ cm, making the whole chord $12$ cm.
$18$ cm, from an arbitrary multiple of $C M$ — Bisection means the two halves are equal, so the chord is exactly double $C M$, not some other multiple.
it cannot be found without knowing both radii — The bisection fact alone gives $C D = 2 \cdot C M = 12$ cm, with no radius needed.
Two circles share centre O, with a chord AB touching the smaller one at P and matching ticks on AP and PB.
Worked example
A chord of the larger circle, bisected where it touches the smaller one
two circles share the same centre $O$; a chord $A B$ of the larger circle touches the smaller circle at the point $P$ This is the given configuration.
join $O P$ This is the auxiliary line the proof turns on.
$O P$ is perpendicular to $A B$ $A B$ is a tangent to the smaller circle at $P$. A radius is always perpendicular to the tangent at its point of contact.
$O P$ is also a line from the centre of the LARGER circle to a point on its own chord $A B$ $O$ is the centre of both circles, and $A B$ is a chord of the larger circle.
a perpendicular from a circle’s centre to a chord always bisects that chord We already know this fact from Class 9.
$A P = B P$ By steps 3 and 4, $O P$ is a perpendicular from the larger circle’s centre to its own chord $A B$. So, by step 5, it bisects that chord.
reflecting the whole figure across the line $O P$ swaps $A$ and $B$ but leaves the picture unchanged Check the result by symmetry. The reflection sends $A P$ to $B P$ and back again, so the two lengths have to be equal.
In proving $A B + C D = A D + B C$ for a quadrilateral circumscribing a circle, what is the key step?
naming the contact points and adding equal tangent pairs
measuring all four sides directly with a ruler
assuming the quadrilateral is a rectangle from the start
applying the RHS congruence criterion directly to the whole quadrilateral
Check your answer
✓ naming the contact points and adding equal tangent pairs — (A) Naming the four points of contact and adding the resulting equal-tangent-length pairs is what produces the side-sum identity.
measuring all four sides directly with a ruler — The proof is a general argument that holds for every such quadrilateral, not a measurement of one particular figure.
assuming the quadrilateral is a rectangle from the start — The identity holds for ANY quadrilateral circumscribing a circle, with no need to assume it is a rectangle.
applying the RHS congruence criterion directly to the whole quadrilateral — RHS congruence compares two triangles; the actual proof instead adds four separate equal-tangent-length pairs.
A quadrilateral $P Q R S$ circumscribes a circle, with $P Q = 7$ cm, $Q R = 5$ cm, $R S = 4$ cm. What is $S P$?
$4$ cm, the same as $R S$
$8$ cm, from a mismatched combination of the given sides
$16$ cm, from adding all three given sides
$6$ cm, from $P Q + R S = Q R + S P$
Check your answer
$4$ cm, the same as $R S$ — Nothing here says opposite sides are equal; the side-sum identity gives $S P = 6$ cm instead.
$8$ cm, from a mismatched combination of the given sides — The identity pairs $P Q$ with $R S$ against $Q R$ with $S P$; mixing the sides differently gives the wrong value.
$16$ cm, from adding all three given sides — Adding all three given sides ignores the side-sum identity entirely, which instead gives $S P = 6$ cm.
✓ $6$ cm, from $P Q + R S = Q R + S P$ — (D) Using the side-sum identity, $7 + 4 = 5 + S P$ gives $S P = 6$ cm.
A student proves $A B + C D = A D + B C$ using only the fact that opposite sides of any quadrilateral are equal. What is wrong with this reasoning?
nothing is wrong, since all quadrilaterals have this property
opposite sides of a general quadrilateral are not equal
the reasoning is wrong because the quadrilateral must be a parallelogram
the reasoning is wrong only when the circle’s radius is very small
Check your answer
nothing is wrong, since all quadrilaterals have this property — Opposite sides are equal only in special quadrilaterals, such as a parallelogram, not in a general one.
✓ opposite sides of a general quadrilateral are not equal — (B) A general quadrilateral has no such opposite-side equality; the true source of the identity is the equal tangent lengths at each vertex.
the reasoning is wrong because the quadrilateral must be a parallelogram — The identity holds for ANY quadrilateral circumscribing a circle; the fix is using the tangent-length argument, not restricting the shape.
the reasoning is wrong only when the circle’s radius is very small — The reasoning is wrong regardless of the radius; opposite sides simply are not equal in a general quadrilateral.
A triangle $A B C$ circumscribes a circle that touches $A B$ at $P$, $B C$ at $Q$, and $C A$ at $R$. Which pairs of segments does the equal-tangent-lengths theorem make equal?
$A P = A R$, $B P = B Q$, and $C Q = C R$
$A P = B Q$ and $B P = C Q$ only
all six tangent segments are equal to each other
$A P = P B$ and $B Q = Q C$
Check your answer
✓ $A P = A R$, $B P = B Q$, and $C Q = C R$ — (A) Each vertex sends out two tangents to its two nearest points of contact, so each such pair is equal.
$A P = B Q$ and $B P = C Q$ only — Equal tangent lengths pair segments from the SAME vertex, such as $A P$ with $A R$, not segments from different vertices.
all six tangent segments are equal to each other — The theorem equates the two tangents from EACH vertex, not all six segments across the whole triangle.
$A P = P B$ and $B Q = Q C$ — $A P$ and $P B$ are tangents from different vertices, $A$ and $B$; the equal pairs instead share a common vertex, like $A P$ and $A R$.
A quadrilateral ABCD surrounds a circle touching all four sides, with matching tick counts of one, two, three and four at each corner.
Worked example
Opposite sides of a quadrilateral that circumscribes a circle
a quadrilateral $A B C D$ is drawn so all four sides touch a circle, at the points $P$, $Q$, $R$, $S$ on sides $A B$, $B C$, $C D$, $D A$ This is the given configuration.
$A P = A S$, $B P = B Q$, $C R = C Q$, $D R = D S$ Tangents drawn from one external point to a circle are always equal in length.
$(A P + P B) + (C R + R D) = (A S + S D) + (B Q + Q C)$ Add the four equal pairs from step 2, grouped by vertex on each side.
$A B + C D = A D + B C$ $A P + P B = A B$, $C R + R D = C D$, $A S + S D = A D$ and $B Q + Q C = B C$. Each pair sits along one side.
for a square of side $s$, each tangent segment is $s/2$, so $A B + C D = 2 s$ and $A D + B C = 2 s$ too Check the identity on a shape you know. A square is a special circumscribing quadrilateral, and both sums come out equal there too.
Why does a parallelogram circumscribing a circle have to be a rhombus?
every shape that circumscribes a circle is automatically a rhombus
the identity plus equal opposite sides forces equal sides
a circle can only be inscribed in a rhombus, never a general parallelogram
the parallelogram’s opposite angles must be right angles
Check your answer
every shape that circumscribes a circle is automatically a rhombus — Plenty of non-parallelogram shapes circumscribe a circle without being a rhombus; the rhombus conclusion needs the parallelogram’s own side equalities too.
✓ the identity plus equal opposite sides forces equal sides — (B) The side-sum identity, combined with the parallelogram’s equal opposite sides, forces every side of the figure to be equal, making it a rhombus.
a circle can only be inscribed in a rhombus, never a general parallelogram — The proof shows a circumscribing PARALLELOGRAM must be a rhombus, not that only rhombi can ever have an inscribed circle drawn near them.
the parallelogram’s opposite angles must be right angles — The proof never needs a right angle anywhere; it uses only the side-sum identity and the parallelogram’s equal opposite sides.
A student argues a RECTANGLE that circumscribes a circle must also be a square, using the same argument as the rhombus proof. Is this valid?
No — the argument only works for non-rectangular parallelograms
Yes — a rectangle is a parallelogram too
No — a circle can never be inscribed in a rectangle
Yes, but only if the rectangle’s diagonals are equal
Check your answer
No — the argument only works for non-rectangular parallelograms — The rhombus argument uses only properties every parallelogram has, a rectangle included, so it applies to a rectangle too.
✓ Yes — a rectangle is a parallelogram too — (B) A rectangle is a special parallelogram, so the same side-sum argument applies and forces all four sides equal, making it a square.
No — a circle can never be inscribed in a rectangle — A circle can be inscribed in a rectangle whose sides happen to satisfy the tangent condition; the question is about what shape it then becomes.
Yes, but only if the rectangle’s diagonals are equal — A rectangle’s diagonals are always equal by definition, so this adds nothing beyond what being a rectangle already guarantees.
Which of these is a direct consequence of the result “a parallelogram circumscribing a circle is a rhombus”?
every rhombus circumscribes some circle
every circle can be circumscribed by some parallelogram
a rhombus always has a larger area than any other parallelogram with the same perimeter
a general, non-rhombus parallelogram can never circumscribe a circle
Check your answer
every rhombus circumscribes some circle — The result proves circumscribing forces a rhombus shape; it does not claim every rhombus has a circle it circumscribes.
every circle can be circumscribed by some parallelogram — This result is about what shape a circumscribing parallelogram must take, not about which parallelograms exist around a given circle.
a rhombus always has a larger area than any other parallelogram with the same perimeter — Area is not part of this result at all; the result is purely about side lengths being forced equal.
✓ a general, non-rhombus parallelogram can never circumscribe a circle — (D) Since circumscribing a circle forces a parallelogram to be a rhombus, any parallelogram that is NOT a rhombus cannot circumscribe one.
A parallelogram ABCD drawn as a diamond around a circle touching all four sides, with one matching tick mark on each side.
Worked example
Why a parallelogram that circumscribes a circle must be a rhombus
a parallelogram $A B C D$ circumscribes a circle This is the given configuration.
$A B = C D$ and $A D = B C$ Opposite sides of a parallelogram are always equal.
$A B + C D = A D + B C$ This is the previous result, since $A B C D$ circumscribes a circle.
$2 \cdot A B = 2 \cdot A D$ Substitute $A B = C D$ and $A D = B C$ from step 2 into step 3.
$A B = A D$ Divide both sides by $2$.
$A B = B C = C D = A D$ Combine $A B = A D$ with the parallelogram’s own $A B = C D$ and $A D = B C$: all four sides come out equal.
$A B C D$ is a rhombus A parallelogram with all four sides equal is a rhombus, by definition.
if $A B = 6$ cm in such a parallelogram, the proof forces $A D = 6$ cm too, so all four sides equal $6$ cm Check the conclusion with a number. Once one side is fixed, the proof pins every other side to the same value, exactly what a rhombus requires.
In proving $\angle P T Q = 2 \cdot \angle O P Q$, why is $\triangle T P Q$ isosceles?
$\angle P T Q$ is given as $90^\circ$
$O$ is equidistant from $T$, $P$, and $Q$
$P$ and $Q$ both lie on the same circle
$T P = T Q$, equal tangents from $T$
Check your answer
$\angle P T Q$ is given as $90^\circ$ — $\angle P T Q$ is not assumed to be any fixed value; the isosceles property comes from the two equal tangent lengths instead.
$O$ is equidistant from $T$, $P$, and $Q$ — $O$ is equidistant from $P$ and $Q$, both radii, but not from $T$, which sits outside the circle at a different distance.
$P$ and $Q$ both lie on the same circle — That $P$ and $Q$ lie on the circle is true but does not by itself make $\triangle T P Q$ isosceles; the equal tangent lengths do.
✓ $T P = T Q$, equal tangents from $T$ — (D) $T P$ and $T Q$ are equal because they are the two tangent lengths from the same external point $T$.
The proof combines the isosceles-triangle angle result with the tangent-perpendicular-to-radius fact. What does the second fact contribute?
it proves $T P = T Q$ directly, without needing the equal-tangent-lengths theorem
it fixes the value of $\angle P T Q$ directly
it fixes $\angle O P T = 90^\circ$, letting $\angle O P Q$ be found by subtraction
it shows that $O$, $P$, and $Q$ are collinear
Check your answer
it proves $T P = T Q$ directly, without needing the equal-tangent-lengths theorem — $T P = T Q$ comes from the equal-tangent-lengths theorem, a separate fact from the tangent-perpendicular-to-radius one.
it fixes the value of $\angle P T Q$ directly — The tangent-radius fact fixes $\angle O P T$, not $\angle P T Q$, which instead is whatever value the problem gives.
✓ it fixes $\angle O P T = 90^\circ$, letting $\angle O P Q$ be found by subtraction — (C) The tangent-radius fact fixes $\angle O P T$ at exactly $90^\circ$, so $\angle O P Q$ can then be found by subtracting $\angle T P Q$ from it.
it shows that $O$, $P$, and $Q$ are collinear — $O$, $P$, and $Q$ form a triangle in this figure, not a straight line; the tangent-radius fact says nothing about collinearity.
Two tangents from an external point $T$ touch a circle at $P$ and $Q$, with $\angle P T Q = 50^\circ$. What is $\angle O P Q$?
$50^\circ$, the same as $\angle P T Q$
$65^\circ$, from an extra subtraction step
$25^\circ$, half of $\angle P T Q$
$100^\circ$, from doubling instead of halving
Check your answer
$50^\circ$, the same as $\angle P T Q$ — $\angle O P Q$ is HALF of $\angle P T Q$, not the same value as it.
$65^\circ$, from an extra subtraction step — The relationship $\angle P T Q = 2 \cdot \angle O P Q$ needs only a halving step, not an additional subtraction.
✓ $25^\circ$, half of $\angle P T Q$ — (C) Since $\angle P T Q$ is twice $\angle O P Q$, halving $50^\circ$ gives $\angle O P Q = 25^\circ$.
$100^\circ$, from doubling instead of halving — $\angle P T Q$ is already the DOUBLE of $\angle O P Q$; finding $\angle O P Q$ needs halving, not a further doubling.
If the two tangents from $T$ were perpendicular to each other, so $\angle P T Q = 90^\circ$, what would $\angle O P Q$ equal?
$90^\circ$, the same as $\angle P T Q$
$0^\circ$, since the tangents already meet at a right angle
it cannot be found without knowing the radius
$45^\circ$, from $\angle P T Q = 2 \cdot \angle O P Q$
Check your answer
$90^\circ$, the same as $\angle P T Q$ — $\angle O P Q$ is HALF of $\angle P T Q$, not equal to it.
$0^\circ$, since the tangents already meet at a right angle — Nothing in the relationship makes $\angle O P Q$ vanish; halving $90^\circ$ gives $45^\circ$, not $0^\circ$.
it cannot be found without knowing the radius — The angle relationship alone is enough here: halving $\angle P T Q$ gives $\angle O P Q$, with no radius needed.
✓ $45^\circ$, from $\angle P T Q = 2 \cdot \angle O P Q$ — (D) Halving the given $90^\circ$ angle, as the relationship requires, gives $\angle O P Q = 45^\circ$.
Two tangents from T touch a circle at P and Q, with angles of 30 degrees at P and 60 degrees at T.
Worked example
Relating the angle between two tangents to the angle at the point of contact
two tangents $T P$ and $T Q$ are drawn from an external point $T$ to a circle with centre $O$, touching it at $P$ and $Q$ This is the given configuration.
$T P = T Q$ The two tangent lengths from one external point are always equal.
$\triangle T P Q$ is isosceles Two of its sides, $T P$ and $T Q$, are equal, from step 2.
$\angle T P Q = 90^\circ - 1/2 \angle P T Q$ The two base angles of an isosceles triangle are equal, and all three angles of a triangle add to $180^\circ$.
$\angle O P T = 90^\circ$ The radius to the point of contact is always perpendicular to the tangent there.
$\angle O P Q = \angle O P T - \angle T P Q$ Angle $O P Q$ is what remains of the right angle at $P$, once angle $T P Q$ is taken out of it.
$\angle O P Q = 90^\circ - (90^\circ - 1/2 \angle P T Q) = 1/2 \angle P T Q$ Substitute step 5 and step 4 into step 6, then simplify.
$\angle P T Q = 2 \cdot \angle O P Q$ Rearrange the previous step.
with $\angle O P Q = 30^\circ$, the relation predicts $\angle P T Q = 2 \cdot 30^\circ = 60^\circ$ Check the relation with the numbers in the figure. A $30^\circ$ angle at $P$ and a $60^\circ$ angle at $T$ fit exactly, since $60$ is double $30$.
This chapter’s results — perpendicularity, equal tangent lengths, and the angle bisector — all trace back to which single fact?
the radius is perpendicular to the tangent there
every chord of a circle is bisected by a radius through its midpoint
all tangents drawn from external points are equal in length
the centre of a circle is equidistant from every point on it
Check your answer
✓ the radius is perpendicular to the tangent there — (A) Every later result in the chapter is built on this one right angle between the radius and the tangent at the point of contact.
every chord of a circle is bisected by a radius through its midpoint — The chord-bisection fact is borrowed once, from Class 9, in a single worked example; it is not the fact this whole chapter is built on.
all tangents drawn from external points are equal in length — Equal tangent lengths is one of the RESULTS proved using the right angle at the point of contact, not the original fact itself.
the centre of a circle is equidistant from every point on it — That the centre is equidistant from every point is simply what a radius is; the chapter’s driving fact is the RIGHT ANGLE the radius makes with the tangent.
If the radius were NOT perpendicular to the tangent at the point of contact, which of this chapter’s results would still stand?
none of them — both later results depend on this same right angle through RHS
the equal-tangent-lengths result, since it only uses the definition of a tangent
the angle-bisector result, since it only uses the equal tangent lengths
all of them, since they are independent facts proved separately
Check your answer
✓ none of them — both later results depend on this same right angle through RHS — (A) Since both the equal-tangent-lengths and angle-bisector results rely on this right angle through RHS congruence, neither would still hold.
the equal-tangent-lengths result, since it only uses the definition of a tangent — The equal-tangent-lengths proof uses RHS congruence, which needs this right angle; without it, that proof no longer goes through.
the angle-bisector result, since it only uses the equal tangent lengths — The angle-bisector result comes from the SAME congruence as the equal-tangent-lengths result, which itself depends on this right angle.
all of them, since they are independent facts proved separately — The results are not independent — both trace back to the same right angle and the same triangle congruence built from it.
The equal-tangent-lengths proof and the angle-bisector fact both use the SAME triangle congruence, $\triangle O Q P \sim \cdot \equiv \triangle O R P$. What does this tell you about the two results?
they are proved completely independently and only coincidentally use the same triangles
one result must be proved before the congruence can be used for the other
they are two different consequences, sides and angles, of one congruence
the two results are actually restatements of the same single fact
Check your answer
they are proved completely independently and only coincidentally use the same triangles — The two results are not independent — both are read off the SAME triangle congruence, one for sides and one for angles.
one result must be proved before the congruence can be used for the other — Once the single congruence is established, both the side result and the angle result follow from it together, with no required order.
✓ they are two different consequences, sides and angles, of one congruence — (C) Both results come from the same congruence: one reads off the equal sides, the other the equal angles, by CPCT.
the two results are actually restatements of the same single fact — Equal tangent lengths and the angle bisector are two DIFFERENT facts, about sides and angles respectively, even though one congruence gives both.
Check yourself: the whole chapter
A line touches a circle at exactly one point, P. Which of these statements about that line must be true?
it must pass through the circle’s centre
it is perpendicular to the radius drawn to P
it must be exactly twice as long as the radius
it cannot be extended beyond the point P in either direction
Check your answer
it must pass through the circle’s centre — The tangent line itself rarely passes through the centre — it is the radius drawn to the point of contact that always does, and that radius is perpendicular to the tangent there.
✓ it is perpendicular to the radius drawn to P — (B) Wherever a line touches a circle at exactly one point, the radius drawn to that point meets the line at a right angle.
it must be exactly twice as long as the radius — Nothing fixes the tangent’s length relative to the radius — a tangent is a full line, not a segment of any particular length.
it cannot be extended beyond the point P in either direction — A tangent is an ordinary straight line extended in both directions — touching the circle at exactly one point does not stop the line there.
In the standard tangent-perpendicular-to-radius proof, $O$ is the centre, $P$ the point of contact, and $Q$ any other point on the tangent. Why is $O Q$ longer than $O P$?
because $Q$ lies inside the circle, and every point inside is closer to $O$ than the circle’s boundary is
because the tangent line is drawn longer than the radius, by convention
because $Q$, being a point on the tangent other than $P$, lies outside the circle, while $P$ lies on it
because $O Q$ and $O P$ are corresponding sides of congruent triangles, which forces them to differ by a fixed ratio
Check your answer
because $Q$ lies inside the circle, and every point inside is closer to $O$ than the circle’s boundary is — A tangent meets the circle at exactly one point, $P$. Any other point on that same line, including $Q$, cannot also be on the circle — it must lie outside.
because the tangent line is drawn longer than the radius, by convention — No such convention exists in the proof. The comparison is between $O Q$ and $O P$ purely because of where $Q$ sits relative to the circle, not because of a length rule for tangent lines.
✓ because $Q$, being a point on the tangent other than $P$, lies outside the circle, while $P$ lies on it — (C) Since the tangent meets the circle only at $P$, any other point $Q$ on it lies outside the circle — so $O Q$, reaching outside, is longer than $O P$, which reaches the boundary exactly.
because $O Q$ and $O P$ are corresponding sides of congruent triangles, which forces them to differ by a fixed ratio — Congruent triangles appear in a different proof in this chapter, the one showing two tangent lengths are equal. This proof compares $O Q$ and $O P$ directly, using no congruence at all.
From an external point $P$, two tangents touch a circle with centre $O$ at $Q$ and $R$. You are told $O Q = O R$ (both radii), $\angle O Q P = \angle O R P = 90^\circ$, and $O P$ is a side shared by both triangles $O Q P$ and $O R P$. What follows about $P Q$ and $P R$?
nothing can be concluded without also knowing the circle’s radius
$P Q$ and $P R$ differ by the length of the chord $Q R$
$P Q$ is greater than $P R$ whenever $Q$ and $R$ lie on opposite sides of the centre
$P Q = P R$, since the two right triangles are congruent by RHS
Check your answer
nothing can be concluded without also knowing the circle’s radius — The three facts given — equal radii, both right angles, and the shared side $O P$ — are exactly the hypotenuse and one pair of sides RHS needs. No numeric radius value is required.
$P Q$ and $P R$ differ by the length of the chord $Q R$ — The chord $Q R$ plays no role in this argument. The equal lengths follow purely from the two triangles being congruent, not from any relationship to the chord joining $Q$ and $R$.
$P Q$ is greater than $P R$ whenever $Q$ and $R$ lie on opposite sides of the centre — Nothing about where $Q$ and $R$ sit changes the result. The same three facts hold regardless of position, and they always give equal triangles, never an inequality.
✓ $P Q = P R$, since the two right triangles are congruent by RHS — (D) A common hypotenuse $O P$, equal radii $O Q = O R$, and a right angle at each of $Q$ and $R$ are exactly what RHS needs — the two triangles are congruent, so $P Q = P R$.
How many tangents can be drawn to a circle from a point that lies INSIDE the circle?
exactly one — this is the count that belongs to a point sitting ON the circle
none — every line through that point is a secant
exactly two — this is the count that belongs to a point OUTSIDE the circle
infinitely many — as if no bound applied to lines through an interior point
Check your answer
exactly one — this is the count that belongs to a point sitting ON the circle — Exactly one tangent belongs to a point ON the circle, at that very point — a point strictly inside the circle is a different case.
✓ none — every line through that point is a secant — (B) Every line through a point inside the circle crosses it twice, making it a secant — no tangent can be drawn from there.
exactly two — this is the count that belongs to a point OUTSIDE the circle — Exactly two tangents belongs to a point OUTSIDE the circle — a point inside the circle has none at all.
infinitely many — as if no bound applied to lines through an interior point — The three cases — inside, on, and outside — give exactly zero, one, and two tangents respectively; none of them is unbounded.
A straight line meets a circle at exactly two points. What is this line called, and what is the single shared point on a TANGENT line called?
the line is a tangent; the single point is called the centre
the line is a chord; the single point is called the radius point
the line is a secant; the single point on a tangent is called the point of contact
the line is a diameter; the single point on a tangent is called the vertex, the term used for a triangle’s corner
Check your answer
the line is a tangent; the single point is called the centre — A tangent meets the circle at exactly one point, never two — that swaps the two definitions. A tangent’s own shared point is called the point of contact, not the centre.
the line is a chord; the single point is called the radius point — A chord is the line segment joining the two points where a secant meets the circle — the secant is the whole line through them, not the segment itself. “Radius point” is not a term this chapter uses.
✓ the line is a secant; the single point on a tangent is called the point of contact — (C) A line meeting a circle at exactly two points is a secant; the one point a tangent shares with the circle is called its point of contact.
the line is a diameter; the single point on a tangent is called the vertex, the term used for a triangle’s corner — A diameter is one special chord, the one passing through the centre — it is not the general term for any line meeting the circle at two points. A tangent’s shared point is its point of contact, not a “vertex”.
TANGENT PERPENDICULAR TO RADIUS: the radius is perpendicular to the tangent at the point of contact. Radius $5$ cm, tangent $12$ cm, angle $90^\circ$.
TANGENT LENGTH BY PYTHAGORAS: $O Q^2 = O P^2 + P Q^2$. For $O P = 5$ cm and $O Q = 13$ cm, $P Q = 12$ cm.
EQUAL TANGENTS: the two tangents from one external point are always equal. Here $P Q = P R = 12$ cm.
ANGLE BISECTOR: $O P$ bisects the angle between the two tangents. A $60^\circ$ angle at $T$ pairs with $30^\circ$ at $P$.
CIRCUMSCRIBING SHAPES: opposite sides of a circumscribing quadrilateral add to the same total. $A B + C D = A D + B C$. Let us look back over the whole chapter and see one fact doing almost all the work. The radius to a point of contact is always perpendicular to the tangent there. That single right angle forces $O Q$ to be longer than $O P$ in the first proof, and it supplies the right angle the RHS criterion needs in the second. Through that same congruence, we get both the equal tangent lengths and the angle bisector at the centre.
A right angle where the radius meets the tangent gives the 5, 12, 13 triangle, and RHS copies the 12 to the other tangent.
Spot the tangent and the right angle comes free, turning a word problem into a Pythagoras sum.
You see round shapes with straight lines near them all the time. Here are seven places a tangent, or its length, decides how something is built or measured.
A bypass road and a roundabout. A bypass road runs past a circular roundabout, touching its edge at exactly one point. The maths: the road that touches the circle once is a tangent. A road that crosses it twice would be a secant.
Laying a path to a fountain. A circular fountain has radius $8$ m. A gate stands $17$ m from the fountain’s centre. A path from the gate touches the fountain’s edge without crossing it. The maths: the path’s length is the tangent length from the gate. It follows from the right angle between the radius and the tangent. $17^2 = 8^2 + \text{path}^2$, so $\text{path}^2 = 289 - 64 = 225$. The path is $15$ m.
Four tangent lengths pair up around the circle, so three edges of the board fix the fourth.
Roping off a circular display. A shop ropes off a round stand, radius $0.9$ m. Two ropes run from a post $1.5$ m from the centre, just touching the edge. The maths: the two tangents drawn from one outside point are always equal in length. Each rope is $\sqrt{1.5^2 - 0.9^2} = \sqrt{1.44} = 1.2$ m, so both ropes measure $1.2$ m.
Drawing a rangoli around a circle. You draw a circle for a rangoli. Then you draw a parallelogram around it, each side just touching the circle. The maths: a parallelogram whose four sides all touch one circle is a rhombus. So all four sides come out equal. If one side is $30$ cm, all four are $30$ cm, and the border is $4 \cdot 30 = 120$ cm long.
Where a running track meets its bend. A running track has straight sections and curved bends. Each straight meets a bend without a kink, touching it at exactly one point. The maths: the straight is a tangent to the bend, and a tangent touches a circle at exactly one point.
Roping off a pond from the bank. You want one straight rope from where you stand to just touch a circular pond, without crossing into the water. The maths: from outside the pond you can draw two such ropes. From a point on the bank, only one. From inside the water, none at all.
Cutting a signboard around a round logo. A four-sided signboard $A B C D$ is cut so each edge just touches a round logo. You know three edges. The maths: when all four sides touch one circle, opposite sides add to the same total. So $A B + C D = A D + B C$. With $A B = 60$, $B C = 40$ and $C D = 50$ cm: $60 + 50 = A D + 40$, so $A D = 70$ cm.
Your turn. You stand $25$ m from the centre of a circular skating rink, radius $7$ m. A straight rope from where you stand just touches the rink’s edge without crossing it. How long is the rope? Answer: $25^2 - 7^2 = 625 - 49 = 576 = 24^2$. The rope is $24$ m.
practice A tangent touches a circle of centre $O$ and radius $3$ cm at the point $P$. An external point $Q$ is $5$ cm from the centre. Find the tangent length $P Q$. (Worked in full below — read it, then do the next two the same way.)
practice A tangent touches a circle of centre $O$ and radius $10$ cm at the point $P$. The tangent length from an external point $Q$ is $P Q = 24$ cm. Find the distance $O Q$. (Same triangle, the other way round — this time the unknown is the hypotenuse, so the two squares are added.)
practice From an external point $Q$, the tangent to a circle with centre $O$ touches it at $P$ and has length $P Q = 20$ cm. The distance $O Q$ is $25$ cm. Find the radius of the circle. (Mark the right angle at $P$ first, then ask which of the three lengths is facing it.)
practice How many tangents can a circle have at one point on it?
practice A circle can have at most how many parallel tangents?
practice A tangent touches a circle of radius $6$ cm at the point $P$. The distance from the centre $O$ to an external point $Q$ is $O Q = 10$ cm. Find the tangent length $P Q$.
practice Draw a circle, then a line lying outside it. Construct two lines parallel to that line — one a tangent to the circle, and the other a secant.
Answers
$P Q^2 = 5^2 - 3^2 = 16$, so $P Q = 4$ cm.
$O Q^2 = 10^2 + 24^2 = 100 + 576 = 676$, so $O Q = 26$ cm.
$O Q$ faces the right angle, so $O P^2 = 25^2 - 20^2 = 625 - 400 = 225$, and the radius is $15$ cm.
Exactly one.
Two — one at each end of a diameter, both perpendicular to it.
$P Q^2 = O Q^2 - O P^2 = 10^2 - 6^2 = 64$, so $P Q = 8$ cm.
Draw the diameter perpendicular to the given line, and place the tangent at the end of that diameter. Then draw any second line parallel to the given line that still cuts the circle in two points — that second line is the secant.
Two panels show compass constructions: one builds a tangent at P using equal arcs, the other draws a line parallel to a given one.
Exercise 10.1 — further practice
practice A line lies in the same plane as a circle and meets it at exactly two points. Name this kind of line, and explain in one sentence how a tangent is different from it.
practice A line and a circle lie in the same plane and do not meet at all. Such a line is:
a tangent to the circle
a secant of the circle
neither a tangent nor a secant
a chord of the circle
practice True or false, with a reason: a line meeting a circle at two points that are extremely close together is a tangent to the circle.
practice A tangent touches a circle of radius $9$ cm at the point $P$. The length of the tangent from an external point $Q$ is $12$ cm. Find the distance $O Q$ from the centre to $Q$.
practice From a point $Q$ outside a circle with centre $O$, the length of the tangent to the circle is $15$ cm, and the distance $O Q$ is $17$ cm. Find the radius of the circle.
practice A tangent touches a circle at the point $P$. A second line is drawn through $P$, perpendicular to this tangent. Prove that this second line passes through the centre of the circle.
practice A tangent to a circle of radius $20$ cm touches it at the point $P$. An external point $Q$ is at a distance $O Q = 29$ cm from the centre. Find the tangent length $P Q$.
practice A circle has radius $7$ cm and centre $O$. The tangent from an external point $Q$ has length $24$ cm. Which of these is the distance $O Q$?
$25$ cm
$26$ cm
$30$ cm
$31$ cm
Answers
This line is a secant. A tangent is different because it meets the circle at exactly one point, not two.
C — neither a tangent nor a secant.
False. However close the two points are, the line stays a secant as long as they remain two distinct points. It becomes a tangent only when the two points coincide exactly into one point.
$O Q = 15$ cm
$O P = 8$ cm
By the point-of-contact theorem, the radius $O P$ is perpendicular to the tangent at $P$. At a given point on a line, only one perpendicular to that line can be drawn, so the second line described is exactly the line $O P$ — which passes through the centre $O$.
practice Two tangents $T P$ and $T Q$ are drawn to a circle with centre $O$ from an external point $T$, touching the circle at $P$ and at $Q$. If $\angle P O Q = 120^\circ$, find $\angle P T Q$ and $\angle O T P$. (Worked in full below — read it, then do the next two the same way.)
practice A quadrilateral $A B C D$ is drawn so that all four of its sides touch a circle. If $A B = 10$ cm, $B C = 8$ cm and $C D = 7$ cm, find $A D$. (Name the four points of contact first. One pair of opposite sides added together equals the other pair added together.)
practice Two circles share the same centre $O$. The larger has radius $17$ cm and the smaller has radius $8$ cm. A chord of the larger circle touches the smaller circle. Find the length of that chord. (The chord is a tangent to the smaller circle, so the radius to its point of contact is perpendicular to it and cuts it in half — find half the chord first, then double it.)
practice Two tangents $T P$ and $T Q$ are drawn to a circle with centre $O$ from an external point $T$. If $\angle P O Q = 110^\circ$, find $\angle P T Q$.
practice Prove that the tangents drawn at the two ends of a diameter of a circle are parallel to each other.
practice Two concentric circles have radii $5$ cm and $r$. A chord of the larger circle, of length $8$ cm, touches the smaller circle. Find $r$.
practice A quadrilateral $A B C D$ circumscribes a circle, touching side $A B$ at $P$, with $A P = 4$ cm and $P B = 3$ cm. If $B C = 9$ cm and $C D = 6$ cm, find $A D$.
practice A triangle $A B C$ circumscribes a circle, touching side $B C$ at $P$, side $C A$ at $Q$, and side $A B$ at $R$. Show that $A R + B P + C Q$ equals half the triangle’s perimeter.
practice Prove that a parallelogram circumscribing a circle is a rhombus.
practice Two tangents $P A$ and $P B$ are drawn from an external point $P$ to a circle with centre $O$. If $\angle A P B = 80^\circ$, find $\angle P O A$.
Answers
In quadrilateral $O P T Q$ the two radii give right angles at $P$ and $Q$, so $\angle P T Q = 360^\circ - 120^\circ - 90^\circ - 90^\circ = 60^\circ$, and since $O T$ bisects it, $\angle O T P = 30^\circ$.
$A B + C D = A D + B C$, so $10 + 7 = A D + 8$ and $A D = 9$ cm.
Half the chord is $\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$ cm, so the whole chord is $30$ cm.
In quadrilateral $O P T Q$, $\angle O P T = \angle O Q T = 90^\circ$ (radius perpendicular to tangent). The four angles sum to $360^\circ$, so $\angle P O Q + \angle P T Q = 180^\circ$, giving $\angle P T Q = 180^\circ - 110^\circ = 70^\circ$.
Let $A B$ be a diameter, with tangents drawn at $A$ and at $B$. Each tangent is perpendicular to the radius at its own point of contact, and both radii lie along the same diameter $A B$. Two lines perpendicular to the same line are parallel, so the two tangents are parallel to each other.
The perpendicular from the common centre to the chord bisects it, giving a half-chord of $4$ cm. In the right triangle this forms, $r^2 + 4^2 = 5^2$, so $r^2 = 9$ and $r = 3$ cm.
$A B = A P + P B = 4 + 3 = 7$ cm. Since $A B + C D = A D + B C$ for any quadrilateral circumscribing a circle, $7 + 6 = A D + 9$, giving $A D = 4$ cm.
Let the tangent lengths from $A$, $B$, $C$ be $x$, $y$, $z$. Then $A R = A Q = x$, $B R = B P = y$, and $C P = C Q = z$, since tangents from one point are always equal. The perimeter is $A B + B C + C A = (x+y)+(y+z)+(z+x) = 2(x+y+z)$. So $A R + B P + C Q = x+y+z$ is exactly half the perimeter.
Let the parallelogram be $A B C D$. Tangents drawn from one point are equal, so adding the two tangent lengths at each vertex around the figure gives $A B + C D = B C + A D$. In a parallelogram $A B = C D$ and $B C = A D$, so $2 A B = 2 B C$, and therefore $A B = B C$. Adjacent sides are equal, so all four sides are equal and the parallelogram is a rhombus.
$O P$ bisects $\angle A P B$, so $\angle A P O = 40^\circ$. Triangle $O A P$ is right-angled at $A$ (radius perpendicular to tangent), so $\angle P O A = 180^\circ - 90^\circ - 40^\circ = 50^\circ$.
A circle with a diameter from A to B, and a tangent line resting on the circle at each of the two ends.
Exercise 10.2 — further practice
practice Two tangents $P A$ and $P B$ are drawn from an external point $P$ to a circle with centre $O$. If $\angle A P B = 64^\circ$, find $\angle A P O$.
practice Two tangents $T P$ and $T Q$ are drawn to a circle with centre $O$ from an external point $T$. If $\angle O P Q = 40^\circ$, find $\angle P T Q$.
practice A parallelogram $P Q R S$ circumscribes a circle. If $P Q = 8$ cm, find $Q R$.
practice Two tangents $T P$ and $T Q$ are drawn to a circle with centre $O$ from an external point $T$. If $\angle P T Q = 54^\circ$, find $\angle O P Q$.
practice A quadrilateral $P Q R S$ circumscribes a circle, touching all four sides. If $P Q = 12$ cm, $Q R = 15$ cm and $R S = 14$ cm, find $S P$.
practice A rectangle $A B C D$ circumscribes a circle, so all four of its sides touch the circle. Prove that $A B C D$ must be a square.
practice A circle is inscribed in a triangle $P Q R$, touching all three sides. The line from the vertex $P$ to the centre $O$ always:
bisects the opposite side $Q R$
bisects the angle at $P$
is perpendicular to $Q R$
is equal in length to the inradius
practice Two concentric circles have centre $O$. The larger circle has radius $13$ cm and the smaller has radius $5$ cm. Find the length of the chord of the larger circle that touches the smaller circle.
practice A triangle $P Q R$ circumscribes a circle, touching side $Q R$ at $S$, side $R P$ at $T$, and side $P Q$ at $U$. If $Q S = 5$ cm, $R S = 4$ cm and $P T = 6$ cm, find the perimeter of triangle $P Q R$.
practice Two tangents $T P$ and $T Q$ are drawn from an external point $T$ to a circle with centre $O$ and radius $5$ cm. If $T O = 13$ cm, find the perimeter of the quadrilateral $O P T Q$.
practice Two parallel tangents to a circle with centre $O$ touch it at the points $M$ and $N$. A third tangent to the same circle, touching it at a point $C$ that lies between the two parallel tangents, meets the tangent at $M$ at a point $A$ and meets the tangent at $N$ at a point $B$, so $A$, $C$ and $B$ are three points on this third tangent line. Prove that $\angle A O B = 90^\circ$.
practice Two tangents $T P$ and $T Q$ are drawn from an external point $T$ to a circle with centre $O$ and radius $9$ cm. The perimeter of quadrilateral $O P T Q$ is $42$ cm. Find $T O$.
Answers
$\angle A P O = 32^\circ$
$\angle P T Q = 80^\circ$
$Q R = 8$ cm
$\angle O P Q = 27^\circ$
$S P = 11$ cm
In a rectangle, opposite sides are equal: $A B = C D$ and $A D = B C$. Since $A B C D$ circumscribes the circle, $A B + C D = A D + B C$. Substituting the equal pairs gives $2 \cdot A B = 2 \cdot A D$, so $A B = A D$. Combined with $A B = C D$ and $A D = B C$, all four sides are equal — and a rectangle with all four sides equal is a square.
B — bisects the angle at $P$.
$A B = 24$ cm
$30$ cm
$34$ cm
$O A$ bisects the angle between the two tangents from $A$, so $\angle O A B = 1/2 \angle M A B$. Likewise $O B$ bisects the angle at $B$, so $\angle O B A = 1/2 \angle N B A$. The tangents at $M$ and $N$ are parallel and $A B$ is a transversal, so $\angle M A B + \angle N B A = 180^\circ$ (co-interior angles). Adding the two bisected halves, $\angle O A B + \angle O B A = 90^\circ$. In triangle $O A B$ the three angles add to $180^\circ$, so $\angle A O B = 90^\circ$.
