MathsCBSE Class 10 · Circles

Tangent Perpendicular to Radius Theorem: Class 10 Proof and Problems

Updated 14 October 2026 · Fact-checked

The theorem says the tangent at any point of a circle makes a right angle with the radius drawn to the point of contact. CBSE asks you to prove it using the shortest distance argument, and to use the right angle it gives in Pythagoras, angle and concentric circle problems worth 1 to 5 marks.

Every other point Q on the tangent lies outside the circle, so OQ is longer than OP; the shortest segment OP must be perpendicular to the tangent.

Illustration drawn for QuestionBuddy.

Understand Tangent Perpendicular to Radius Theorem

A tangent is a line that meets a circle at exactly one point, called the point of contact. Theorem 10.1 tells you how the tangent sits relative to the centre: the tangent at any point of a circle is perpendicular to the radius through the point of contact. If the tangent $X Y$ touches the circle with centre $O$ at $P$ , then $O P ⊥ X Y$ .

Why is this true? Take any point $Q$ on $X Y$ other than $P$ . The tangent meets the circle only at $P$ , so $Q$ must lie outside the circle. Every point outside a circle is farther from the centre than the radius, so $O Q > O P$ . This holds for every such $Q$ . So $O P$ is the shortest of all segments from $O$ to the line $X Y$ . Of all segments from a point to a line, the shortest one is the perpendicular. Hence $O P ⊥ X Y$ .

The theorem gives you a free right angle at every point of contact. Join the centre $O$ to an external point $P$ and to the point of contact $T$ : $△ O T P$ is right-angled at $T$ with hypotenuse $O P$ , so $P T^{2} = O P^{2} - r^{2}$ . The right angle also lets you use trigonometric ratios, angle sums in triangles, and the fact that an isosceles triangle $O A B$ (two radii) has equal base angles.

Several results follow straight from it. The line through the point of contact perpendicular to the tangent passes through the centre, because there is only one perpendicular to a line at a given point. Tangents at the two ends of a diameter are both perpendicular to that diameter, so they are parallel. In concentric circles, a chord of the larger circle that touches the smaller one is perpendicular to the small radius at the contact point, so it is bisected there. Theorem 10.2 (equal tangents) also uses this right angle in its RHS congruence.

The conditions matter. The right angle is between the tangent and the radius at the point of contact only. A radius drawn to any other point of the circle is not perpendicular to the tangent, and a line from the centre to an external point is not a radius. In every answer, write the reason for the right angle as "tangent is perpendicular to the radius at the point of contact" before you use it.

Tangent Perpendicular to Radius Theorem: formulas to remember

Theorem 10.1: $O P ⊥ X Y, ∠ O P X = ∠ O P Y = 9 0^{\circ}$; P is the point of contact of tangent XY. Quote this as the reason for every right angle at a point of contact.
Tangent length by Pythagoras: $P T^{2} = O P^{2} - r^{2}$; Right triangle OTP with the right angle at T. OP, from the centre to the external point, is always the hypotenuse.
Trigonometry in the tangent triangle: $sin ∠ O P T = \frac{r}{O P}, tan ∠ O P T = \frac{r}{P T}$; Use when an angle at the external point is given.
Angle between tangent and a chord from the point of contact: $∠ O A B = 9 0^{\circ} - ∠ B A T$; AT is the tangent at A and AB a chord. Then use OA = OB to get the other angles of triangle OAB.
Chord of the larger of two concentric circles touching the smaller: $A B = 2 R^{2} - r^{2}$; The chord is perpendicular to the small radius at the contact point, so it is bisected there.
Tangents at the ends of a diameter: $l ∥ m, distance between them = 2 r$; Both are perpendicular to the same diameter.

How to solve Tangent Perpendicular to Radius Theorem questions

Almost every question on this theorem becomes a right-triangle question once you mark the right angle. Use this method each time.

1Describe or sketch the figure: centre $O$ , point of contact, tangent line, and any external point.
2Join the centre to the point of contact. Mark $9 0^{\circ}$ there and write the reason: tangent is perpendicular to the radius at the point of contact.
3Identify the right triangle. The side from the centre to a point on the tangent (not the contact point) is the hypotenuse.
4Convert disguised data: distance from an external point to the nearest point of the circle plus $r$ gives $O P$ ; a diameter gives $2 r$ .
5Choose the tool: Pythagoras for lengths, trigonometric ratios if an angle is given, angle sum and isosceles triangles (two radii) for angles.
6For proofs, write Given, To prove, Construction and Proof, with a reason in brackets on every line.
7Check the answer: the hypotenuse must be the longest side, and angles in each triangle must add to $18 0^{\circ}$ .

Tangent Perpendicular to Radius Theorem in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tangent perpendicular to radius

Suppose a circular park has a fountain at its centre O and radius 40 m. A straight road touches the edge of the park at T. A bus stop S is on the road, 30 m from T.

(i) Find the distance from the bus stop to the fountain. (ii) A girl walks from S straight towards the fountain. How far does she walk before she reaches the edge of the park? (iii) The town wants a second stop on the road, on the other side of T, that is also as far from the fountain as S is. How far apart will the two stops be?

Try it first, then show the solutionSolution

The road is a tangent at T, so $O T ⊥$ road and $△ O T S$ is right-angled at T.
(i) $O S = 4 0^{2} + 3 0^{2} = 2500 = 50$ m.
(ii) Walking along SO, she reaches the edge after $50 - 40 = 10$ m.
(iii) A point R on the road with $O R = 50$ m has $T R = 5 0^{2} - 4 0^{2} = 30$ m, so R is 30 m from T on the other side.
The two stops are $30 + 30 = 60$ m apart.

Answer: (i) 50 m (ii) 10 m (iii) 60 m apart.

Memory hook: Road touches the park: radius to the touch point stands square to the road.

Tricks to remember Tangent Perpendicular to Radius Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Shortest is straight
Every other point of the tangent is outside, so it is farther away. The shortest path from the centre to the tangent is the radius, and the shortest path to a line is always the perpendicular.
Use it when: Recalling the logic of the proof of Theorem 10.1 in the right order.
Radius meets tangent, think 90
Whenever a radius ends where a tangent touches, draw a small square at that corner before doing anything else.
Use it when: Spotting the right triangle in length and angle problems.
Hyp is the far one
In triangle OTP, the side that does not touch T is the hypotenuse: OP, centre to external point.
Use it when: Avoiding writing OP squared equal to PT squared plus OT squared the wrong way round.

Tangent Perpendicular to Radius Theorem: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Tangent length from an angle · 2 marks · Very short answerBoard level

$P T$ is a tangent to a circle with centre $O$ , touching it at $T$ . If $O P = 12$ cm and $∠ O P T = 3 0^{\circ}$ , find the radius of the circle and the length $P T$ .

Try it first, then show the solutionSolution

$∠ O T P = 9 0^{\circ}$ (tangent is perpendicular to the radius at the point of contact). (½ mark)
In right $△ O T P$ : $sin 3 0^{\circ} = \frac{O T}{O P}$ , so $O T = 12 \times \frac{1}{2} = 6$ cm. (½ mark)
$cos 3 0^{\circ} = \frac{P T}{O P}$ , so $P T = 12 \times \frac{3}{2} = 63$ cm. (1 mark)
Check: $6^{2} + (63)^{2} = 36 + 108 = 144 = 1 2^{2}$ .

Answer: Radius $= 6$ cm, $P T = 63$ cm.

Example 2: Concentric circles with an unknown pair of radii · 3 marks · Short answerHarder board level

Two concentric circles have centre $O$ . A chord $A B$ of the larger circle touches the smaller circle at $P$ , and $A B = 16$ cm. The radius of the larger circle is $2$ cm more than the radius of the smaller circle. Find both radii.

Try it first, then show the solutionSolution

$O P ⊥ A B$ (tangent to the smaller circle is perpendicular to the radius $O P$ at the point of contact). (½ mark)
The perpendicular from the centre to a chord bisects it, so $A P = 8$ cm. (½ mark)
Let the radii be $R$ and $r$ . In right $△ O P A$ : $R^{2} - r^{2} = 8^{2} = 64$ . (½ mark)
$(R - r) (R + r) = 64$ and $R - r = 2$ , so $R + r = 32$ . (½ mark)
Adding: $2 R = 34$ , so $R = 17$ cm and $r = 15$ cm. (1 mark)
Check: $1 7^{2} - 1 5^{2} = 289 - 225 = 64$ .

Answer: Larger radius $17$ cm, smaller radius $15$ cm.

Example 3: Angle between tangent and chord · 3 marks · Short answerBoard level

$A T$ is the tangent at $A$ to a circle with centre $O$ , and $A B$ is a chord. If $∠ B A T = 5 5^{\circ}$ , find $∠ O A B$ , $∠ O B A$ and $∠ A O B$ . Give reasons.

Try it first, then show the solutionSolution

$∠ O A T = 9 0^{\circ}$ (tangent is perpendicular to the radius at the point of contact). (1 mark)
$∠ O A B = 9 0^{\circ} - 5 5^{\circ} = 3 5^{\circ}$ . (½ mark)
$O A = O B$ (radii), so $∠ O B A = ∠ O A B = 3 5^{\circ}$ (angles opposite equal sides). (½ mark)
$∠ A O B = 18 0^{\circ} - 3 5^{\circ} - 3 5^{\circ} = 11 0^{\circ}$ (angle sum of a triangle). (1 mark)

Answer: $∠ O A B = 3 5^{\circ}$ , $∠ O B A = 3 5^{\circ}$ , $∠ A O B = 11 0^{\circ}$ .

Example 4: Proof of Theorem 10.1 with an application · 5 marks · Long answerHarder board level

(i) Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
(ii) $P Q$ is the tangent at $P$ to a circle with centre $O$ and radius $9$ cm. $Q$ is a point on the tangent such that the segment $O Q$ meets the circle at $R$ with $QR = 32$ cm. Find $P Q$ .

Try it first, then show the solutionSolution

Given: a circle with centre $O$ and a tangent $X Y$ touching it at $P$ . To prove: $O P ⊥ X Y$ . (½ mark)
Construction: take any point $Q$ on $X Y$ other than $P$ and join $O Q$ . (½ mark)
$Q$ lies outside the circle, because a tangent meets the circle at only one point, $P$ . (½ mark)
So $O Q$ is longer than the radius: $O Q > O P$ . This holds for every point $Q$ of $X Y$ other than $P$ . (½ mark)
Hence $O P$ is the shortest segment from $O$ to $X Y$ , and the shortest segment from a point to a line is the perpendicular. So $O P ⊥ X Y$ . (1 mark)
(ii) $O Q = O R + R Q = 9 + 32 = 41$ cm. (½ mark)
$∠ O P Q = 9 0^{\circ}$ (by part (i)). (½ mark)
$P Q^{2} = O Q^{2} - O P^{2} = 1681 - 81 = 1600$ , so $P Q = 40$ cm. (1 mark)

Answer: (i) Proved. (ii) $P Q = 40$ cm.

Example 5: Case study: ruler touching a round table top · 4 marks · Case-basedHOTS / competency

In a class activity, a circular table top has centre $O$ and radius $60$ cm. A long straight ruler is laid flat so that it just touches the edge of the table at $T$ . An ant starts at a point $P$ on the ruler, $80$ cm from $T$ , and walks along the ruler towards $T$ .
(i) Why is $O T$ perpendicular to the ruler? (1 mark)
(ii) How far is the ant from $O$ at the start? (1 mark)
(iii) At which point of its walk is the ant closest to $O$ , and how far has it walked when it is first $65$ cm from $O$ ? (2 marks)

Try it first, then show the solutionSolution

(i) The ruler touches the circle at only $T$ , so it is a tangent, and the tangent is perpendicular to the radius $O T$ at the point of contact. (1 mark)
(ii) $O P^{2} = O T^{2} + T P^{2} = 3600 + 6400 = 10000$ , so $O P = 100$ cm. (1 mark)
(iii) $O T$ is the shortest distance from $O$ to the ruler, so the ant is closest to $O$ at $T$ , at a distance of $60$ cm. (½ mark)
Let $Q$ be the point where $O Q = 65$ cm. $T Q^{2} = 6 5^{2} - 6 0^{2} = 4225 - 3600 = 625$ , so $T Q = 25$ cm. (1 mark)
Distance walked $= P T - T Q = 80 - 25 = 55$ cm. (½ mark)

Answer: (i) Tangent is perpendicular to radius at the point of contact. (ii) $100$ cm. (iii) Closest at $T$ ( $60$ cm); it has walked $55$ cm.

Example 6: Tangents at the ends of a diameter · 2 marks · Very short answerBoard level

AB is a diameter of a circle with centre O. Lines l and m are the tangents at A and B. Prove that $l ∥ m$ .

Try it first, then show the solutionSolution

The tangent is perpendicular to the radius at the point of contact, so $l ⊥ O A$ and $m ⊥ O B$ , that is, both $l$ and $m$ are perpendicular to the line AB. (1 mark)
With AB as a transversal, the co-interior angles at A and B are $9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ , so $l ∥ m$ . (1 mark)

Answer: Both tangents are perpendicular to the diameter AB, so they are parallel.

Where students lose marks in Tangent Perpendicular to Radius Theorem

The slips examiners see most often, and what to write instead.

Mistake: Using the right angle without a reason
Examiners award the reason mark separately; a bare $9 0^{\circ}$ looks assumed.
Wrong working
$∠ O T P = 9 0^{\circ}$ , so $P T^{2} = O P^{2} - O T^{2}$ .
Right working
$∠ O T P = 9 0^{\circ}$ (tangent is perpendicular to the radius at the point of contact), so $P T^{2} = O P^{2} - O T^{2}$ .
Do this: Write the reason in brackets the first time you use the right angle.
Usually costs: ½ to 1 mark
Mistake: Taking the wrong side as hypotenuse
Students add instead of subtract because they treat $P T$ as the longest side.
Wrong working
$O P = 13$ , $r = 5$ : $P T^{2} = 1 3^{2} + 5^{2} = 194$ .
Right working
$P T^{2} = O P^{2} - r^{2} = 169 - 25 = 144$ , so $P T = 12$ .
Do this: The hypotenuse is opposite the right angle at $T$ : always $O P$ .
Usually costs: 1 to 2 marks
Mistake: Skipping why $Q$ is outside the circle in the proof
The whole proof depends on $O Q > O P$ , which needs this reason.
Wrong working
$O Q > O P$ , so $O P$ is shortest, hence $O P ⊥ X Y$ .
Right working
$Q$ lies outside the circle since the tangent meets it only at $P$ ; so $O Q > O P$ for every such $Q$ . The shortest segment to a line is the perpendicular, so $O P ⊥ X Y$ .
Do this: Learn the proof as a chain of three reasons: outside, longer, shortest is perpendicular.
Usually costs: 1 mark
Mistake: Using the distance to the circle as $O P$
The question gives the gap from $P$ to the circle, not to the centre.
Wrong working
$P$ is $8$ cm from the circle, $r = 6$ : $P T^{2} = 64 - 36 = 28$ .
Right working
$O P = 8 + 6 = 14$ , so $P T^{2} = 196 - 36 = 160$ , $P T = 410$ cm.
Do this: Add the radius to any distance measured to the circle.
Usually costs: 1 to 2 marks

Exam tips

In proof questions, write Given, To prove, Construction and Proof as separate headings and describe the figure in words.
Name the theorem as the reason (tangent is perpendicular to radius at the point of contact) every time you mark a right angle.
When an angle at the external point is given, think trigonometry in the right triangle; when lengths are given, think Pythagoras.
In concentric circle questions, state both reasons: perpendicular by the theorem, and bisected because a perpendicular from the centre bisects a chord.
Check every length answer: the hypotenuse $O P$ must come out longer than both the radius and the tangent.

Practice questions on Tangent Perpendicular to Radius Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A tangent $P T$ from a point $P$ touches a circle with centre $O$ and radius $8$ cm at $T$ . If $P T = 15$ cm, the distance from $P$ to the nearest point of the circle is
(a) $17$ cm
(b) $9$ cm
(c) $7$ cm
(d) $23$ cm
Hint
Find $O P$ first using the right angle at $T$ , then subtract the radius.
Answer
(b) $9$ cm. $O P = 64 + 225 = 17$ cm, nearest point is $17 - 8 = 9$ cm away.
Q2 · 3 marks · Short answerHarder board level
$A B$ is a diameter of a circle with centre $O$ , $A B = 6$ cm. The tangent at $B$ has a point $C$ on it with $B C = 8$ cm. $A C$ meets the circle again at $D$ . Find $A C$ and $B D$ .
Hint
$∠ A B C = 9 0^{\circ}$ by the theorem; $∠ A D B = 9 0^{\circ}$ as an angle in a semicircle. Use area of $△ A B C$ two ways.
Answer
$A C = 36 + 64 = 10$ cm. Area $= \frac{1}{2} (6) (8) = \frac{1}{2} (10) (B D)$ , so $B D = 4.8$ cm.
Q3 · 3 marks · Short answerBoard level
$P T$ is a tangent to a circle with centre $O$ and radius $6$ cm, touching it at $T$ , and $O P = 12$ cm. Find $∠ T P O$ , $∠ T O P$ and the area of $△ O T P$ .
Hint
In right $△ O T P$ , $sin ∠ T P O = \frac{O T}{O P}$ .
Answer
$sin ∠ T P O = \frac{1}{2}$ , so $∠ T P O = 3 0^{\circ}$ , $∠ T O P = 6 0^{\circ}$ . $P T = 144 - 36 = 63$ cm. Area $= \frac{1}{2} (6) (63) = 183$ cm².
Q4 · 5 marks · Long answerHarder board level
Two concentric circles with centre $O$ have radii $10$ cm and $6$ cm. Chord $A B$ of the larger circle touches the smaller circle at $P$ .
(i) Prove that $P$ is the midpoint of $A B$ and find $A B$ .
(ii) The tangents to the larger circle at $A$ and $B$ meet at $T$ . Find $O T$ and $T A$ .
Hint
$O P ⊥ A B$ by the theorem. For (ii), $O$ , $P$ , $T$ are collinear and $△ O A T$ is right-angled at $A$ with $A P$ as altitude, so $△ O P A \sim △ O A T$ .
Answer
(i) $O P ⊥ A B$ , and the perpendicular from the centre bisects the chord; $A P = 100 - 36 = 8$ , $A B = 16$ cm. (ii) From similarity, $O A^{2} = O P \cdot O T$ , so $O T = \frac{100}{6} = \frac{50}{3}$ cm; $T A = \frac{2500}{9} - 100 = \frac{40}{3}$ cm.
Q5 · 3 marks · Short answerHOTS / competency
A student writes: "The tangent at $P$ touches a circle of radius $5$ cm with centre $O$ . $Q$ is a point on the tangent with $O Q = 4$ cm." (i) Explain why this is impossible. (ii) If instead $O Q = 13$ cm, find $P Q$ . (iii) What is the least possible value of $O Q$ for a point $Q$ on the tangent?
Hint
Every point of a tangent other than the contact point is outside the circle.
Answer
(i) $Q$ on the tangent is outside or at $P$ , so $O Q \geq 5$ cm; $4$ cm is impossible. (ii) $P Q = 169 - 25 = 12$ cm. (iii) $5$ cm, when $Q = P$ .
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): No tangent can be drawn to a circle from a point inside it.
Reason (R): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is R the reason why every line through an inside point cuts the circle twice?
Answer
(b). A is true: every line through an inside point cuts the circle at two points, so none is a tangent. R is the true theorem but it does not explain A.
Q7 · 4 marks · Long answerHarder board level
Tangents TP and TQ are drawn from an external point T to a circle with centre O and radius 5 cm, and $O T = 10$ cm. Find $∠ O T P$ , $∠ P T Q$ and $∠ P O Q$ , the length PT, the area of quadrilateral OPTQ, and the length of chord PQ.
Hint
In right triangle OPT, $sin ∠ O T P = \frac{O P}{O T}$ .
Answer
$∠ O P T = 9 0^{\circ}$ , so $sin ∠ O T P = \frac{5}{10} = \frac{1}{2}$ and $∠ O T P = 3 0^{\circ}$ . OT bisects $∠ P T Q$ , so $∠ P T Q = 6 0^{\circ}$ and $∠ P O Q = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ . $P T = 100 - 25 = 53$ cm. Area of OPTQ $= 2 \times \frac{1}{2} \times 5 \times 53 = 253 \approx 43.3$ cm $^{2}$ . $T P = T Q$ and $∠ P T Q = 6 0^{\circ}$ , so $△ P T Q$ is equilateral and $P Q = 53$ cm.
Q8 · 3 marks · Short answerHarder board level
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Hint
Join the ends of the chord to the centre and use the right angles at the points of contact.
Answer
Let AB be the chord, O the centre, and AT, BT the tangents at A and B. $O A = O B$ (radii), so $∠ O A B = ∠ O B A$ . $∠ O A T = ∠ O B T = 9 0^{\circ}$ (tangent ⟂ radius). So $∠ T A B = 9 0^{\circ} - ∠ O A B = 9 0^{\circ} - ∠ O B A = ∠ T B A$ : the tangents make equal angles with the chord.
Q9 · 4 marks · Case-basedHOTS / competency
Case study: A wheel of radius 30 cm stands upright on level ground, touching it at T. A straight pole is fixed perpendicular to the ground at T.
(i) Explain why the pole passes through the centre O of the wheel. (1 mark)
(ii) A point P on the ground is 80 cm from T. Find the distance from P to the top of the wheel (the point of the wheel on the pole farthest from T). (1 mark)
(iii) Find the distance from P to the nearest point of the wheel. (2 marks)
Hint
The ground is a tangent at T; the perpendicular to a tangent at the point of contact passes through the centre.
Answer
(i) The ground is a tangent to the wheel at T, and the radius OT is perpendicular to it. The perpendicular to a tangent at the point of contact is unique, so the pole lies along OT and passes through O. (1)
(ii) The top of the wheel is 60 cm above T on the pole, so the distance is $8 0^{2} + 6 0^{2} = 100$ cm. (1)
(iii) $O P = 8 0^{2} + 3 0^{2} = 7300 = 1073 \approx 85.44$ cm. The nearest point of the wheel lies on OP, so the distance is $85.44 - 30 \approx 55.44$ cm. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Theorem 10.1.Show answer
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2.In right triangle OTP formed by a tangent, which side is the hypotenuse?Show answer
$O P$ , from the centre to the external point.
3.Why is a point $Q$ on a tangent, other than the contact point, outside the circle?Show answer
Because a tangent meets the circle at only one point.
4.Tangents at the ends of a diameter are related how?Show answer
They are parallel, both being perpendicular to the diameter.
5.Length of a chord of a circle of radius $R$ touching a concentric circle of radius $r$ ?Show answer
$2 R^{2} - r^{2}$

Tangent Perpendicular to Radius Theorem: frequently asked questions

Do I need to learn the proof of the tangent perpendicular to radius theorem?

Yes. The syllabus lists it as a theorem to prove, so it can be asked as a full proof or as part of a longer question. Learn the shortest distance argument with all its reasons.

Is the theorem true for a radius drawn to any point of the circle?

No. The right angle is only between the tangent and the radius at the point of contact. A radius to any other point is not perpendicular to that tangent.

What is the converse of this theorem?

The line drawn through the point of contact perpendicular to the tangent passes through the centre. It follows because there is only one perpendicular to a line at a given point.

Are constructions of tangents in the 2026-27 syllabus?

No. Constructions of tangents were removed and are not in the 2026-27 syllabus.