Exercise 10.2 — Circles

Applying the two theorems — tangent lengths, MCQs with justification, and circumscribing shapes

13 questions✓ Free · step-by-step

12 marksNCERT Cl-10 Maths, Ex 10.2, Q1
From a point Q, the length of the tangent to a circle is 24 cm, and the distance of Q from the centre is 25 cm. The radius of the circle is: (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm.
Hint. The tangent, the radius to the point of contact, and the line from Q to the centre form a right triangle — with the distance to the centre as the hypotenuse.
Step 1 — Identify the right triangle. Let the point of contact be P. Since OP ⊥ (tangent), △OPQ is right-angled at P, with OQ = 25 as the hypotenuse and PQ = 24 as one leg.

Step 2 — Apply Pythagoras. OQ² = OP² + PQ² 25² = OP² + 24² 625 = OP² + 576 OP² = 49

Step 3 — Take the root. OP = 7

✦ Answer: (A) 7 cm

Notice this is the (7, 24, 25) Pythagorean triple — recognising it would have skipped the arithmetic entirely.
Where students slip. Adding instead of subtracting: 24² + 25² gives a much larger, wrong number. The tangent length and the radius are the two legs, and the distance to the centre is the hypotenuse — so the hypotenuse squared is what you subtract from.
Another way. Spot (7, 24, 25) as a standard triple straightaway, the same way (5, 12, 13) and (8, 15, 17) come up elsewhere in this book.
22 marksNCERT Cl-10 Maths, Ex 10.2, Q2 (Fig. 10.11)
TP and TQ are tangents from an external point T to a circle with centre O, and ∠POQ = 110°. Then ∠PTQ is equal to: (A) 60° (B) 70° (C) 80° (D) 90°.
Hint. OPTQ is a quadrilateral, and two of its angles are right angles by Theorem 10.1. The four angles must sum to 360°.
Step 1 — Mark the right angles. P and Q are the points of contact, so by Theorem 10.1, ∠OPT = ∠OQT = 90°.

Step 2 — Use the angle sum of the quadrilateral OPTQ. ∠O + ∠P + ∠T + ∠Q = 360° 110° + 90° + ∠PTQ + 90° = 360°

Step 3 — Solve. ∠PTQ = 360° − 110° − 90° − 90° = 70°

✦ Answer: (B) 70°

Worth remembering as a general pattern: ∠PTQ and ∠POQ always add to 180°, since 90° + 90° already accounts for half the quadrilateral's total.
Where students slip. Treating OPTQ as if its angles summed to 180°, as in a triangle. It has four vertices, so the angle sum is 360°, not 180°.
Another way. Use the shortcut directly: ∠PTQ = 180° − ∠POQ = 180° − 110° = 70°. This 'supplementary' relationship is exactly what Exercise 10.2 Q10 asks you to prove in general.
32 marksNCERT Cl-10 Maths, Ex 10.2, Q3
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to: (A) 50° (B) 60° (C) 70° (D) 80°.
Hint. OP bisects the angle between the two tangents. That gives you one angle of the right triangle OAP directly.
Step 1 — Split the angle at P. By Theorem 10.2's remark, OP bisects ∠APB. So ∠OPA = 80°/2 = 40°

Step 2 — Use the right angle at A. A is the point of contact, so ∠OAP = 90° by Theorem 10.1.

Step 3 — Use the angle sum of △OAP. ∠OAP + ∠OPA + ∠POA = 180° 90° + 40° + ∠POA = 180°

Step 4 — Solve. ∠POA = 180° − 90° − 40° = 50°, because those are the only three angles the triangle has and they must add to 180°.

✦ Answer: (A) 50°
Where students slip. Using the full 80° in triangle OAP instead of the bisected 40°. The 80° is the angle between the *two* tangents at P; each half-triangle only sees half of it.
Another way. Since ∠OAP = 90° always, ∠POA and ∠OPA are complementary in this right triangle — so ∠POA = 90° − 40° = 50° in one step, once the bisection is done.
43 marksNCERT Cl-10 Maths, Ex 10.2, Q4
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Hint. Both tangents are perpendicular to the same line — the diameter itself. Two lines perpendicular to the same line are parallel to each other.
Step 1 — Name the tangents. Let AB be a diameter of a circle with centre O, and let ℓ and m be the tangents at A and B respectively.

Step 2 — Apply Theorem 10.1 at each end. OA is the radius to the point of contact A, so ℓ ⊥ OA, i.e. ℓ ⊥ AB. OB is the radius to the point of contact B, so m ⊥ OB, i.e. m ⊥ AB.

Step 3 — Both tangents are perpendicular to the same line AB. Two lines that are each perpendicular to a common line are parallel to one another.

✦ Answer: ℓ ∥ m — proved.

The diameter is what makes this work: A, O and B are collinear, so 'radius at A' and 'radius at B' are actually the same straight line, giving both tangents a shared perpendicular.
Where students slip. Trying to prove this for two tangents at the ends of an arbitrary chord. It is false in general — the argument depends entirely on A, O, B being collinear, which only happens when AB is a diameter.
Another way. Coordinates make it visual: place O at the origin with A = (−r, 0) and B = (r, 0). The tangent at A is the vertical line x = −r and at B is x = r — manifestly parallel.
53 marksNCERT Cl-10 Maths, Ex 10.2, Q5
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Hint. This is the converse direction of Theorem 10.1. Suppose the perpendicular does not pass through the centre, and see what goes wrong.
Step 1 — Describe the two lines at P. Let a circle have centre O, and let AB be a tangent touching it at P. Let PQ be the line through P perpendicular to AB. We must show PQ passes through O.

Step 2 — Recall Theorem 10.1. The tangent AB is perpendicular to the radius OP, i.e. OP ⊥ AB.

Step 3 — Compare the two perpendiculars. At the single point P, there is only one line perpendicular to AB — you cannot draw two different perpendiculars to the same line from the same point. Since both OP and PQ are perpendicular to AB at P, they must be the same line.

Step 4 — Conclude. So PQ is the same line as OP, and since OP passes through the centre O, so does PQ.

✦ Answer: the perpendicular to the tangent at the point of contact passes through the centre — proved.

This is really Theorem 10.1 read backwards: that theorem says the radius is perpendicular to the tangent; this one says any perpendicular to the tangent at that point must therefore be the radius.
Where students slip. Trying to prove this by constructing a fresh perpendicular and a separate argument, without noticing that the uniqueness of the perpendicular at a point is the entire proof.
Another way. Argue by contradiction explicitly: if the perpendicular PQ did not pass through O, you would have two different perpendiculars to AB through the same point P, which is impossible in a plane.
62 marksNCERT Cl-10 Maths, Ex 10.2, Q6
The length of a tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Hint. The distance to the centre is the hypotenuse of the tangent right triangle.
Step 1 — Draw the right triangle formed by the radius, the tangent, and the line to the external point. Let the point of contact be P. Then OP ⊥ AP, so △OPA is right-angled at P, with OA = 5 as the hypotenuse and AP = 4 as one leg.

Step 2 — Apply Pythagoras. OA² = OP² + AP² 25 = OP² + 16 OP² = 9

Step 3 — Take the root. OP = 3

✦ Answer: the radius is 3 cm, because that is what OP represents in this triangle.

This is the (3, 4, 5) triple — the smallest and most recognisable one.
Where students slip. Confusing which quantity is the hypotenuse. The distance from the external point to the *centre* is always the hypotenuse; the tangent length and the radius are the two legs.
Another way. Recognise (3, 4, 5) immediately: with two of the three numbers given as 5 and 4, the third must be 3.
73 marksNCERT Cl-10 Maths, Ex 10.2, Q7
Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Hint. This chord is tangent to the smaller circle, so the radius of the smaller circle to the point of contact is perpendicular to it — and that same point bisects the chord.
Step 1 — Name the circles and the chord. Let O be the common centre, and let AB be a chord of the larger circle (radius 5) that touches the smaller circle (radius 3) at point P.

Step 2 — Use the tangency, which is why OP is perpendicular to the chord. AB is tangent to the smaller circle at P, so by Theorem 10.1, OP ⊥ AB, and OP = 3 (the smaller radius).

Step 3 — Use the perpendicular-bisects-chord fact. AB is also a chord of the larger circle, and the perpendicular from a circle's centre to a chord bisects that chord (a Class 9 result). So P is the midpoint of AB, and OA = 5 is the larger radius.

Step 4 — Apply Pythagoras in right triangle OPA. OA² = OP² + PA² 25 = 9 + PA² PA² = 16, so PA = 4

Step 5 — Double it for the full chord. AB = 2 × PA = 8

✦ Answer: the chord is 8 cm long.

This is exactly the situation the chapter's own Example 1 sets up in general — this question is the numeric version of it.
Where students slip. Reporting PA = 4 as the final answer. That is only half the chord — from the centre's foot to one end — the full chord AB is twice that.
Another way. The (3, 4, 5) triple appears again: with radii 3 and 5, the half-chord is automatically 4, and doubling gives 8 without needing to write out Pythagoras.
84 marksNCERT Cl-10 Maths, Ex 10.2, Q8 (Fig. 10.12)
A quadrilateral ABCD is drawn to circumscribe a circle, touching the sides at P, Q, R, S. Prove that AB + CD = AD + BC.
Hint. Name the tangent length from each vertex with its own letter. Every side is then a sum of two of those four letters.
Step 1 — Name the four tangent lengths. Let the circle touch AB at P, BC at Q, CD at R, and DA at S. By Theorem 10.2, the two tangents from each vertex are equal:

AP = AS = w BP = BQ = x CQ = CR = y DR = DS = z

Step 2 — Write each side as a sum of two tangent lengths. AB = AP + PB = w + x BC = BQ + QC = x + y CD = CR + RD = y + z DA = DS + SA = z + w

Step 3 — Add opposite sides. AB + CD = (w + x) + (y + z) = w + x + y + z AD + BC = (z + w) + (x + y) = w + x + y + z

Step 4 — Compare. Both sums equal the same total, w + x + y + z, since addition doesn't care about the order the four letters are grouped in, so

AB + CD = AD + BC

✦ Answer: Proved.
Where students slip. Assigning a separate, unrelated variable to each of the eight tangent segments (AP, PB, BQ, ... individually) instead of recognising that Theorem 10.2 pairs them into just four unknowns. That makes the algebra far longer than it needs to be.
Another way. Notice the pattern once proved: for *any* quadrilateral circumscribing a circle, the sum of one pair of opposite sides always equals the sum of the other pair. This exact fact is what Exercise 10.2 Q11 uses to show a circumscribing parallelogram must be a rhombus.
94 marksNCERT Cl-10 Maths, Ex 10.2, Q9 (Fig. 10.13)
XY and X′Y′ are two parallel tangents to a circle with centre O, and another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Hint. OA bisects the angle at A between its two tangents, and OB bisects the angle at B. The two parallel tangents fix what those two angles add to.
Step 1 — Identify the tangents from A and from B. From A, the two tangents to the circle are AX (part of line XY) and AC (part of AB). By the angle-bisector remark under Theorem 10.2, AO bisects ∠XAC (called ∠A here for short).

Similarly, from B the two tangents are BY′ (part of X′Y′) and BC (part of AB), and BO bisects ∠Y′BC (called ∠B).

Step 2 — Use the parallel tangents. XY ∥ X′Y′ with the transversal AB crossing both. So the co-interior angles at A and B (the ones on the same side, between the two parallels) add up to 180°:

∠A + ∠B = 180°

Step 3 — Take half of each. ∠OAB = ∠A/2 and ∠OBA = ∠B/2, so

∠OAB + ∠OBA = (∠A + ∠B)/2 = 180°/2 = 90°

Step 4 — Use the angle sum of △OAB. ∠OAB + ∠OBA + ∠AOB = 180° 90° + ∠AOB = 180° ∠AOB = 90°

✦ Answer: Proved.
Where students slip. Assuming ∠A = ∠B by symmetry without justification. They need not be equal individually — only their *sum* is fixed by the parallel lines, which is all the proof actually uses.
Another way. Since ∠OAB + ∠OBA = 90° exactly (from Step 3), you can also see immediately that △OAB must be right-angled at O without going through the general angle-sum step separately — the two base angles already used up 90°, leaving 90° for the vertex.
103 marksNCERT Cl-10 Maths, Ex 10.2, Q10
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Hint. This is exactly the general version of Exercise 10.2 Q2. Use the quadrilateral formed by the centre, the two points of contact, and the external point.
Step 1 — Name the points involved. Let P be the external point, PA and PB the two tangents touching the circle at A and B, and O the centre. We must show ∠APB + ∠AOB = 180°.

Step 2 — Mark the right angles. A and B are points of contact, so by Theorem 10.1, ∠OAP = ∠OBP = 90°.

Step 3 — Use the angle sum of quadrilateral OAPB. ∠OAP + ∠APB + ∠PBO + ∠BOA = 360° 90° + ∠APB + 90° + ∠AOB = 360°

Step 4 — Simplify. ∠APB + ∠AOB = 360° − 180° = 180°

✦ Answer: ∠APB and ∠AOB are supplementary — proved.

This is precisely the relationship used numerically in Exercise 10.2 Q2 (∠PTQ = 70° when ∠POQ = 110°, and indeed 70° + 110° = 180°).
Where students slip. Re-deriving this from scratch for each specific numeric question rather than recognising it as one general fact worth remembering: the two right angles from Theorem 10.1 always account for 180° of the quadrilateral's 360°, leaving the other two angles to share the remaining 180° between them.
Another way. State it as a one-line consequence: since OAPB always has two right angles at A and B, the remaining angles at O and P must together make up 360° − 180° = 180°, regardless of the specific circle or point chosen.
113 marksNCERT Cl-10 Maths, Ex 10.2, Q11
Prove that the parallelogram circumscribing a circle is a rhombus.
Hint. A parallelogram already has equal opposite sides. Combine that with the equal-sum property from Exercise 10.2 Q8 to force all four sides equal.
Step 1 — Use the circumscribing property. Let ABCD be a parallelogram circumscribing a circle. By the result of Exercise 10.2 Q8 (equal tangent lengths from each vertex), any quadrilateral circumscribing a circle satisfies

AB + CD = AD + BC … (1)

Step 2 — Use the parallelogram property. In a parallelogram, opposite sides are equal:

AB = CD and AD = BC … (2)

Step 3 — Substitute (2) into (1). AB + AB = AD + AD 2·AB = 2·AD AB = AD

Step 4 — Combine with (2). AB = AD, and AD = BC, and BC = AB (going around using (2) again), so

AB = BC = CD = DA

All four sides are equal, which is why the shape qualifies as more than just any parallelogram.

✦ Answer: a parallelogram with all four sides equal is, by definition, a rhombus — proved.
Where students slip. Trying to show the diagonals are perpendicular, or some other rhombus property, from scratch. The definition of a rhombus is simply 'a parallelogram with all sides equal' — once all four sides are shown equal, nothing else needs proving.
Another way. This result explains something practical: if you are ever told a circle is inscribed in a parallelogram, you may immediately treat every side as equal — a fact several harder problems assume without restating it.
125 marksNCERT Cl-10 Maths, Ex 10.2, Q12 (Fig. 10.14)
A triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC into which the point of contact D divides BC are 8 cm and 6 cm respectively. Find the sides AB and AC.
Hint. Name the unknown tangent length from A as x. Write the area of the triangle two different ways — once from Heron's formula, once as r times the semi-perimeter — and equate.
Step 1 — Name the tangent lengths. Let the circle touch BC at D, CA at E, AB at F. By Theorem 10.2: BD = BF = 8, CD = CE = 6, and let AF = AE = x (unknown).

So AB = AF + FB = x + 8, and AC = AE + EC = x + 6, and BC = 8 + 6 = 14.

Step 2 — Write the semi-perimeter. s = (AB + BC + CA)/2 = ((x+8) + 14 + (x+6))/2 = (2x + 28)/2 = x + 14

Step 3 — Write the area using the incircle radius. Area = r × s = 4(x + 14)

Step 4 — Write the area using Heron's formula. s − a = (x+14) − 14 = x (a = BC) s − b = (x+14) − (x+6) = 8 (b = CA) s − c = (x+14) − (x+8) = 6 (c = AB)

Area² = s(s−a)(s−b)(s−c) = (x+14)(x)(8)(6) = 48x(x+14)

Step 5 — Equate the two expressions for area, squaring the one from Step 3. [4(x+14)]² = 48x(x+14) 16(x+14)² = 48x(x+14)

Divide both sides by 16(x+14), valid since x + 14 ≠ 0: (x+14) = 3x 14 = 2x x = 7

Step 6 — Find the sides. AB = x + 8 = 15 AC = x + 6 = 13

Check. s = 21, Area by Heron = √(21 × 7 × 8 × 6) = √7056 = 84. Area by r·s = 4 × 21 = 84 ✓

✦ Answer: AB = 15 cm and AC = 13 cm.
Where students slip. Trying to find AB and AC using Pythagoras, as if the triangle were right-angled. Nothing in the question says ABC has a right angle — the correct route uses the incircle's area relationship, not Pythagoras.
Another way. Once x = 7 is found, AB, BC, CA = 15, 14, 13 is a well-known triangle with integer sides and area 84 — recognising these numbers is a quick way to check the arithmetic.
134 marksNCERT Cl-10 Maths, Ex 10.2, Q13
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Hint. Join the centre to all four points of contact and to all four vertices. Look for pairs of congruent triangles created by the equal tangent lengths.
Step 1 — Draw in all the radii and vertex-to-centre lines. Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S respectively. Join OA, OB, OC, OD, OP, OQ, OR, OS. We must show ∠AOB + ∠COD = 180° (and consequently ∠BOC + ∠DOA = 180° too).

Step 2 — Find congruent triangle pairs using equal tangents. AP = AS (tangents from A), and OA is common, and OP = OS (radii). So △OAP ≅ △OAS by SSS, giving ∠AOP = ∠AOS = a (say)

Similarly: ∠BOP = ∠BOQ = b, ∠COQ = ∠COR = c, ∠DOR = ∠DOS = d

Step 3 — Use the fact that the eight angles around O add to 360°. 2a + 2b + 2c + 2d = 360° a + b + c + d = 180°

Step 4 — Express the two angles we want in terms of these. ∠AOB = ∠AOP + ∠POB = a + b ∠COD = ∠COR + ∠ROD = c + d

Step 5 — Add them. ∠AOB + ∠COD = (a + b) + (c + d) = a + b + c + d = 180°, since that sum was already fixed in Step 3.

✦ Answer: ∠AOB + ∠COD = 180° — proved. By the same argument, ∠BOC + ∠DOA = 180° as well, so both pairs of opposite sides subtend supplementary angles at the centre.
Where students slip. Assuming the four angles a, b, c, d are all equal to each other. They need not be — a circumscribing quadrilateral need not be regular in any way. Only their total, 180°, is forced.
Another way. This result is the circle-geometry twin of Exercise 10.2 Q9: there, two parallel tangents plus a third gave a 90° angle at the centre; here, all four sides of a general circumscribing quadrilateral give two supplementary pairs at the centre. Seeing both as 'the angle bisectors from each vertex meet at O and share the 360° around it' unifies the two proofs.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter is unchanged by rationalisation — two exercises, 10.1 with 4 questions and 10.2 with 13, and the same two theorems in the summary). Questions are referenced from the NCERT textbook for identification.

All exercises in Circles

Exercise 10.1 (4 Q)Exercise 10.2 (13 Q)

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