NCERT Solutions for Class 10 Maths Chapter 10: Circles (NCERT 2026–27)

These Class 10 Maths Chapter 10 solutions cover Circles from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 10.1 and Exercise 10.2 is reproduced exactly as in the book and solved step by step, with clear reasoning on tangents, the radius–tangent perpendicularity, and the equal-tangents theorem — so you can master the chapter and revise it quickly before exams.

Class: 10 Subject: Mathematics Chapter: 10 — Circles Exercises: 10.1 (4 Q), 10.2 (13 Q) Topic: Tangents to a circle Session: 2026–27
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Chapter 10 Overview

Chapter 10 of Class 10 Maths, Circles, studies how a line and a circle meet in a plane. A line may be non-intersecting (no common point), a secant (two common points), or a tangent (exactly one common point, called the point of contact). The chapter proves two key results — the tangent at any point of a circle is perpendicular to the radius through the point of contact (Theorem 10.1), and the lengths of the two tangents drawn from an external point are equal (Theorem 10.2). These ideas are then applied to chords, concentric circles, and figures circumscribing a circle, such as quadrilaterals and triangles. The Class 10 Maths Chapter 10 solutions below work through every question of Exercise 10.1 and Exercise 10.2 step by step.

Key Concepts & Definitions

Tangent to a circle: a line that touches the circle at exactly one point. That point is the point of contact.

Secant: a line that intersects a circle in two distinct points. A tangent is the limiting case of a secant when its two intersection points coincide.

Non-intersecting line: a line that has no common point with the circle.

Length of the tangent: the length of the segment from an external point to the point of contact on the circle.

Number of tangents: from a point inside a circle – none; from a point on the circle – exactly one; from a point outside the circle – exactly two.

Normal: the line containing the radius through the point of contact is called the normal to the circle at that point.

Important Formulas & Theorems (Chapter 10)

Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact. So if PT is a tangent at P and O is the centre, then OP ⊥ PT.

Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. If PQ and PR are tangents from P, then PQ = PR.

Length of tangent: from an external point at distance d from the centre of a circle of radius r, the tangent length is √(d2 − r2) (right-angled triangle, by Pythagoras).

Angle property: the centre lies on the bisector of the angle between two tangents from an external point (OP bisects ∠QPR).

Circumscribing a circle: for a quadrilateral whose sides all touch a circle, AB + CD = AD + BC (sum of opposite sides are equal).

Exercise 10.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified.

1. How many tangents can a circle have?

SOLUTION At every single point on a circle there is exactly one tangent, and a circle has infinitely many points on it. ∴ A circle can have infinitely many tangents.

2. Fill in the blanks : (i) A tangent to a circle intersects it in ________ point (s). (ii) A line intersecting a circle in two points is called a ________. (iii) A circle can have ________ parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called ________.

SOLUTION (i) A tangent touches the circle at one point → one point(s). (ii) A line meeting the circle in two points is a secant. (iii) Tangents parallel to a given direction occur on the two opposite sides of the circle → two parallel tangents at the most. (iv) The single common point of a tangent and the circle is the point of contact.

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm   (B) 13 cm   (C) 8.5 cm   (D) √119 cm.

SOLUTION PQ is a tangent at P, so by Theorem 10.1 the radius OP ⊥ PQ. Hence triangle OPQ is right-angled at P. By Pythagoras: OQ2 = OP2 + PQ2, so PQ2 = OQ2 − OP2 = 122 − 52 = 144 − 25 = 119. ∴ PQ = √119 cm — the correct option is (D) √119 cm.

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

SOLUTION (construction described in words) Step 1: Draw a circle with centre O and any convenient radius. Step 2: Draw a given line l somewhere outside or across the circle (this is the reference line whose direction we must match). Step 3: Draw a line m parallel to l that just touches the circle at a single point — this line is the required tangent (the radius to the point of contact is perpendicular to m). Step 4: Draw another line n parallel to l that cuts the circle at two points — this line is the required secant. Thus m (tangent) and n (secant) are both parallel to the given line l, as required.

Exercise 10.2 Solutions

In Q.1 to 3, choose the correct option and give justification.

1. 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

SOLUTION Let the tangent touch the circle at P. Then OP (radius) ⊥ QP, so triangle OPQ is right-angled at P. OQ2 = OP2 + QP2 ⇒ OP2 = OQ2 − QP2 = 252 − 242 = 625 − 576 = 49. ∴ radius OP = √49 = 7 cm — the correct option is (A) 7 cm.

2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to (A) 60°   (B) 70°   (C) 80°   (D) 90°

SOLUTION TP and TQ are tangents at P and Q, so by Theorem 10.1, ∠OPT = 90° and ∠OQT = 90°. In quadrilateral OPTQ, the angle sum is 360°: ∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°. 110° + 90° + ∠PTQ + 90° = 360° ⇒ ∠PTQ = 360° − 290° = 70°. ∴ ∠PTQ = (B) 70°.

3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to (A) 50°   (B) 60°   (C) 70°   (D) 80°

SOLUTION Tangents PA and PB give ∠APB = 80°. By symmetry, OP bisects ∠APB, so ∠OPA = 40°. OA is the radius to the point of contact A, so ∠OAP = 90° (Theorem 10.1). In triangle OAP: ∠POA = 180° − ∠OAP − ∠OPA = 180° − 90° − 40° = 50°. ∴ ∠POA = (A) 50°.

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

SOLUTION Let AB be a diameter of a circle with centre O. Let l be the tangent at A and m be the tangent at B. By Theorem 10.1, the tangent at A is perpendicular to radius OA, so l ⊥ AB. Similarly m ⊥ AB. Now AB is a transversal cutting l and m. The angles ∠(l, AB) = 90° and ∠(m, AB) = 90° are equal alternate angles (each 90°). Two lines both perpendicular to the same line AB are parallel to each other. ∴ tangent l ∥ tangent m, i.e. the tangents at the ends of a diameter are parallel. Hence proved.

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

SOLUTION Let a line touch the circle at point P, and let O be the centre. Draw the perpendicular to the tangent at P; call it the line PN. By Theorem 10.1, the radius OP is perpendicular to the tangent at the point of contact P. At the point P there can be one and only one line perpendicular to the tangent. Since both OP and PN are perpendicular to the tangent at P, they must be the same line. As OP passes through the centre O, the line PN (the perpendicular at the point of contact) also passes through O. ∴ the perpendicular at the point of contact to a tangent passes through the centre. Hence proved.

6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

SOLUTION Let the tangent touch the circle at P. Then OP ⊥ AP (Theorem 10.1), so triangle OPA is right-angled at P. Here OA = 5 cm (distance from centre) and AP = 4 cm (tangent length). By Pythagoras: OA2 = OP2 + AP2 ⇒ OP2 = 52 − 42 = 25 − 16 = 9. ∴ radius OP = √9 = 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

SOLUTION Let O be the common centre. Let AB be a chord of the larger circle (radius 5 cm) that touches the smaller circle (radius 3 cm) at point P. AB is a tangent to the smaller circle at P, so OP ⊥ AB and OP = 3 cm. The perpendicular from the centre to a chord bisects it, so AP = PB. In right triangle OPA: OA = 5 cm (radius of larger circle), OP = 3 cm. AP2 = OA2 − OP2 = 25 − 9 = 16, so AP = 4 cm. ∴ chord AB = 2 × AP = 2 × 4 = 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

SOLUTION Let the circle touch sides AB, BC, CD and DA at points P, Q, R and S respectively. Tangents from an external point are equal (Theorem 10.2). So from each vertex: From A: AP = AS  •  From B: BP = BQ  •  From C: CR = CQ  •  From D: DR = DS. Add these: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ). i.e. AB + CD = AD + BC, since AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC. ∴ AB + CD = AD + BC. Hence proved.

9. In 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 intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

SOLUTION Let the tangent XY touch the circle at P and X′Y′ touch it at Q; AB touches at C. So A is external to two tangents AP and AC, and B is external to two tangents BQ and BC. Tangents from an external point are equally inclined to the line joining that point to the centre, so OA bisects ∠PAC and OB bisects ∠QBC. Since XY ∥ X′Y′ and AB is a transversal, the co-interior angles add to 180°: ∠PAB + ∠QBA = 180°, i.e. ∠PAC + ∠QBC = 180°. Then ∠OAC + ∠OBC = ½∠PAC + ½∠QBC = ½(180°) = 90°. In triangle AOB: ∠AOB = 180° − (∠OAC + ∠OBC) = 180° − 90° = 90°. ∴ ∠AOB = 90°. Hence proved.

10. 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.

SOLUTION Let PA and PB be the two tangents from external point P, touching the circle (centre O) at A and B. We must prove ∠APB + ∠AOB = 180°. OA ⊥ PA and OB ⊥ PB (Theorem 10.1), so ∠OAP = 90° and ∠OBP = 90°. In quadrilateral OAPB, the four angles sum to 360°: ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°. 90° + ∠APB + 90° + ∠AOB = 360° ⇒ ∠APB + ∠AOB = 180°. ∴ the angle between the tangents is supplementary to the angle ∠AOB subtended at the centre by AB. Hence proved.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

SOLUTION Let ABCD be a parallelogram that circumscribes a circle. As proved in Q.8, for any quadrilateral circumscribing a circle, AB + CD = AD + BC. In a parallelogram, opposite sides are equal: AB = CD and AD = BC. Substitute into AB + CD = AD + BC: AB + AB = AD + AD, i.e. 2AB = 2AD, so AB = AD. Then AB = CD = AD = BC, so all four sides are equal. A parallelogram with all sides equal is a rhombus. Hence proved.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

SOLUTION Let the incircle (radius r = 4 cm, centre O) touch BC at D, CA at E and AB at F. By equal tangents (Theorem 10.2): BD = BF = 8 cm, CD = CE = 6 cm, and let AF = AE = x. Then AB = AF + FB = (x + 8), AC = AE + EC = (x + 6), BC = BD + DC = 8 + 6 = 14 cm. Sides: a = BC = 14, b = CA = x + 6, c = AB = x + 8. Semi-perimeter s = (14 + x + 6 + x + 8)/2 = (2x + 28)/2 = x + 14. Area by Heron’s formula: s − a = x, s − b = 8, s − c = 6, so Area = √[(x + 14)(x)(8)(6)] = √[48x(x + 14)]. Also Area = r × s = 4(x + 14). Equate: 4(x + 14) = √[48x(x + 14)]. Square both sides: 16(x + 14)2 = 48x(x + 14) ⇒ 16(x + 14) = 48x ⇒ x + 14 = 3x ⇒ 2x = 14 ⇒ x = 7. ∴ AB = x + 8 = 15 cm and AC = x + 6 = 13 cm.

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

SOLUTION Let quadrilateral ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S respectively. We prove ∠AOB + ∠COD = 180° (and similarly ∠BOC + ∠DOA = 180°). Join O to A, B, C, D and to the points of contact. Two tangents from one vertex are equally inclined to the line from that vertex to O, so they subtend equal angles at O. Let: ∠1 = ∠AOP = ∠AOS, ∠2 = ∠BOP = ∠BOQ, ∠3 = ∠COQ = ∠COR, ∠4 = ∠DOR = ∠DOS (equal pairs at O). All eight angles around O add to 360°: 2(∠1 + ∠2 + ∠3 + ∠4) = 360°, so ∠1 + ∠2 + ∠3 + ∠4 = 180°. Now ∠AOB = ∠1 + ∠2 and ∠COD = ∠3 + ∠4. Adding: ∠AOB + ∠COD = (∠1 + ∠2 + ∠3 + ∠4) = 180°. ∴ ∠AOB + ∠COD = 180°, and likewise ∠BOC + ∠AOD = 360° − 180° = 180°. Opposite sides subtend supplementary angles at the centre. Hence proved.

Common Mistakes to Avoid

Watch out for these

  • Forgetting that the radius to the point of contact is perpendicular to the tangent — this right angle is the key to almost every numerical in this chapter.
  • Mixing up the distance from the centre (the hypotenuse) with the tangent length (a leg) in the Pythagoras step — the centre–point distance is always the longest side.
  • In quadrilateral-angle proofs, using a triangle angle sum (180°) when the figure has four vertices — a quadrilateral’s angles add to 360°.
  • Assuming a parallelogram circumscribing a circle is a rectangle — it is a rhombus (equal sides), not necessarily right-angled.
  • In the “circumscribing quadrilateral” result, forgetting to pair tangents correctly from each vertex before adding.
  • Treating √119 as a value that simplifies — 119 = 7 × 17 has no square factor, so leave it as √119.

Practice MCQs & Assertion–Reason

1. A tangent to a circle touches it in exactly:

(a) 0 points    (b) 1 point    (c) 2 points    (d) infinitely many points

2. The angle between a tangent to a circle and the radius at the point of contact is:

(a) 45°    (b) 60°    (c) 90°    (d) 180°

3. The number of tangents that can be drawn from a point inside a circle is:

(a) 0    (b) 1    (c) 2    (d) infinite

4. The number of tangents that can be drawn from a point outside a circle is:

(a) 0    (b) 1    (c) 2    (d) 3

5. The length of the tangent from a point 13 cm from the centre of a circle of radius 5 cm is:

(a) 8 cm    (b) 12 cm    (c) 18 cm    (d) √194 cm

6. If two tangents from an external point are inclined at 60°, the angle subtended by the chord of contact at the centre is:

(a) 60°    (b) 90°    (c) 120°    (d) 150°

7. A circle can have how many parallel tangents at most in any one direction?

(a) 1    (b) 2    (c) 3    (d) infinite

8. For a quadrilateral ABCD circumscribing a circle, which relation always holds?

(a) AB + BC = CD + DA    (b) AB + CD = AD + BC    (c) AB = CD    (d) AC = BD

9. A parallelogram that circumscribes a circle is always a:

(a) rectangle    (b) square    (c) rhombus    (d) trapezium

10. Two tangents TP and TQ are drawn to a circle with centre O. If ∠POQ = 120°, then ∠PTQ equals:

(a) 50°    (b) 60°    (c) 70°    (d) 90°

Answer key: 1-(b), 2-(c), 3-(a), 4-(c), 5-(b), 6-(c), 7-(b), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The tangent length from a point 25 cm from the centre of a circle of radius 7 cm is 24 cm.

Reason: The radius to the point of contact is perpendicular to the tangent, so the tangent length is √(d2 − r2).

A-R 2. Assertion: The lengths of the two tangents drawn from an external point to a circle are equal.

Reason: The two right triangles formed are congruent by the RHS rule.

A-R 3. Assertion: From a point inside a circle, exactly two tangents can be drawn.

Reason: Every line through a point inside a circle meets the circle in two points.

A-R 4. Assertion: The tangents at the two ends of a diameter of a circle are parallel.

Reason: Both tangents are perpendicular to the same diameter, and two lines perpendicular to the same line are parallel.

A-R 5. Assertion: A parallelogram circumscribing a circle is a square.

Reason: In a quadrilateral circumscribing a circle, the sums of opposite sides are equal.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(D).

Quick Revision Summary

  • A tangent touches a circle at exactly one point (the point of contact); a secant cuts it in two points.
  • The tangent at any point is perpendicular to the radius through the point of contact (Theorem 10.1).
  • From an external point exactly two tangents can be drawn, and their lengths are equal (Theorem 10.2).
  • Tangent length from distance d to a circle of radius r = √(d2 − r2).
  • The angle between two tangents and the angle subtended by their chord of contact at the centre are supplementary.
  • For a quadrilateral circumscribing a circle, AB + CD = AD + BC; a circumscribing parallelogram is a rhombus.
  • Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.

How to score full marks in this chapter

Start every tangent numerical by marking the right angle between the radius and the tangent, then write the Pythagoras relation OQ2 = OP2 + PQ2 clearly. For proofs, state the theorem you use (10.1 or 10.2) before applying it, and quote the angle-sum of the correct polygon (180° for a triangle, 360° for a quadrilateral). In figure questions, name the points of contact and label equal tangents from each vertex — that single labelling earns most of the proof marks.

Frequently Asked Questions

What is Class 10 Maths Chapter 10 Circles about?

Chapter 10, Circles, studies tangents to a circle: how a line can be a tangent, secant or non-intersecting line, the proof that a tangent is perpendicular to the radius at the point of contact, and the proof that the two tangents from an external point are equal in length, with applications to chords and circumscribing figures.

How many exercises are there in Class 10 Maths Chapter 10?

There are two exercises — Exercise 10.1 with 4 questions and Exercise 10.2 with 13 questions — all solved step by step on this page for the NCERT 2026–27 textbook.

What is the tangent length from an external point?

If an external point is at distance d from the centre of a circle of radius r, the length of the tangent is √(d2 − r2), because the radius, the tangent and the line to the centre form a right-angled triangle.

Are these Class 10 Maths Chapter 10 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 10 Maths Chapter 10 Circles are free and follow the official NCERT textbook for the 2026–27 session, with every answer verified.

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