CBSE Class 10 Mathematics: NCERT Solutions Class 10 Maths Chapter 10 Circles (2025–2026 Guide)

Mastering CBSE Class 10 Mathematics requires clarity on geometric proofs, and these complete NCERT Solutions Class 10 Maths Chapter 10 Circles are designed to help you score 100% in your 2025–2026 board exam. In the CBSE Class 10 syllabus, Chapter 10 is one of the most scoring units in geometry, carrying significant weight in short-answer and long-answer sections. This comprehensive guide covers fundamental definitions, formal theorem proofs, step-by-step textbook solutions, and high-frequency questions for board exam 10 preparation.

Key Concepts and Geometric Definitions

Before solving exercise problems in CBSE Mathematics, it is crucial to clearly distinguish how lines interact with circles in a coordinate or Euclidean plane.

• Non-intersecting Line: A line that has no common point with the circle. The perpendicular distance from the centre of the circle to the line is strictly greater than the radius.
• Secant: A straight line that intersects the circle at two distinct points.
• Tangent: A straight line that touches the circle at exactly one point. The shared point between the circle and the line is known as the point of contact.

Key Tangent Properties to Remember

• There is only one tangent passing through a given point lying on a circle.
• There are zero tangents from a point located strictly inside a circle.
• There are exactly two tangents that can be drawn from an external point to a circle.
• A circle can have a maximum of two parallel tangents at any given time, which occur at the opposite endpoints of a diameter.

Core Theorems and Step-by-Step Proofs

Class 10 Circles contains two foundational theorems. Direct proof questions on these theorems appear frequently in Section C and Section D of the CBSE Class 10 Mathematics board papers.

Theorem 10.1: Radius Perpendicular to Tangent

Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a tangent line XY touching the circle at point P.

To Prove: OP ⊥ XY.

Proof:

1. Take any point Q on the tangent line XY other than point P, and join OQ.
2. Point Q must lie strictly outside the circle. (If Q lay inside the circle, the line XY would intersect the circle at two points and become a secant, contradicting the fact that XY is a tangent).
3. Because Q lies outside the circle, the length of segment OQ is greater than the radius OP:
   OQ > OP.
4. Since this inequality holds true for every point on line XY except the point of contact P, OP is the shortest distance from the centre O to the line XY.
5. In Euclidean geometry, the shortest segment connecting a point to a line is the perpendicular segment.
6. Therefore, OP ⊥ XY. Hence proved.

Theorem 10.2: Lengths of Tangents from an External Point

Statement: The lengths of tangents drawn from an external point to a circle are equal.

Given: A circle with centre O and an external point P. Two tangents PQ and PR are drawn from P touching the circle at points Q and R respectively.

To Prove: PQ = PR.

Construction: Join OP, OQ, and OR.

Proof:

1. By Theorem 10.1, the radius drawn to the point of contact is perpendicular to the tangent. Thus, ∠OQP = 90° and ∠ORP = 90°.
2. Consider right-angled triangles ΔOQP and ΔORP:
   • ∠OQP = ∠ORP = 90° (Radii perpendicular to tangents at contact points)
   • OQ = OR (Radii of the same circle)
   • OP = OP (Common hypotenuse)
3. By RHS (Right angle-Hypotenuse-Side) congruence criterion:
   ΔOQP ≅ ΔORP.
4. Corresponding parts of congruent triangles are equal (CPCT). Therefore, PQ = PR. Hence proved.

Key Corollaries from Theorem 10.2:

• ∠OPQ = ∠OPR: The line joining the external point to the centre bisects the angle between the two tangents.
• ∠POQ = ∠POR: The tangents subtend equal angles at the centre of the circle.

NCERT Solutions and Important CBSE Board Questions

Question 1: Tangent Length Calculation

Problem: 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 such that OQ = 12 cm. Find the length of PQ.

Solution:

1. By Theorem 10.1, the radius OP is perpendicular to tangent PQ at point P. Therefore, ΔOPQ is a right-angled triangle with ∠OPQ = 90°.
2. Applying the Pythagoras theorem in ΔOPQ:
   OQ² = OP² + PQ²
3. Substitute the given dimensions:
   12² = 5² + PQ²
   144 = 25 + PQ²
   PQ² = 144 − 25 = 119
   PQ = √119 cm.

Question 2: Angle Subtended by Tangents

Problem: If tangents PA and PB from an external point P to a circle with centre O are inclined to each other at an angle of 80°, find ∠POA.

Solution:

1. From Theorem 10.2, the line segment OP bisects ∠APB.
   ∠APO = ½ × ∠APB = ½ × 80° = 40°.
2. By Theorem 10.1, the radius OA is perpendicular to the tangent PA, so ∠OAP = 90°.
3. In ΔOAP, the sum of interior angles is 180°:
   ∠POA + ∠OAP + ∠APO = 180°
   ∠POA + 90° + 40° = 180°
   ∠POA + 130° = 180°
   ∠POA = 50°.

Question 3: Parallel Tangents at Endpoints of Diameter

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

Solution:

1. Let AB be a diameter of a circle with centre O. Let line l be the tangent at point A and line m be the tangent at point B.
2. By Theorem 10.1, radius OA ⊥ line l ⇒ ∠OAP = 90° (where P is a point on l).
3. Similarly, radius OB ⊥ line m ⇒ ∠OBQ = 90° (where Q is a point on m on the alternate side).
4. Since AOB is a straight line, ∠OAP and ∠OBQ form a pair of alternate interior angles for lines l and m with transversal AB.
5. Because ∠OAP = ∠OBQ = 90°, alternate interior angles are equal.
6. Therefore, tangent line l is parallel to line m. Hence proved.

Question 4: Parallelogram Circumscribing a Circle

Problem: Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

1. Let ABCD be a parallelogram circumscribing a circle touching sides AB, BC, CD, and DA at points P, Q, R, and S respectively.
2. Using Theorem 10.2 (tangents from an external point are equal):
   • AP = AS (Tangents from A) — (Equation 1)
   • BP = BQ (Tangents from B) — (Equation 2)
   • CR = CQ (Tangents from C) — (Equation 3)
   • DR = DS (Tangents from D) — (Equation 4)
3. Adding Equations (1), (2), (3), and (4):
   (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
   AB + CD = AD + BC — (Equation 5)
4. Since ABCD is a parallelogram, opposite sides are equal: AB = CD and AD = BC.
5. Substituting these into Equation (5):
   AB + AB = AD + AD
   2AB = 2AD ⇒ AB = AD.
6. A parallelogram with adjacent sides equal is a rhombus. Thus, ABCD is a rhombus. Hence proved.

Question 5: Circumscribed Triangle Side Lengths

Problem: A triangle ABC circumscribes a circle of radius 4 cm such that segments BD and DC formed by contact point D on BC are 8 cm and 6 cm respectively. Find sides AB and AC.

Solution:

1. Let the circle touch AC at E and AB at F.
2. By Theorem 10.2:
   • CF = CD = 6 cm
   • BF = BD = 8 cm
   • Let AE = AF = x cm
3. Side lengths: a = BC = 14 cm, b = AC = (6 + x) cm, c = AB = (8 + x) cm.
4. Semi-perimeter s = [14 + (6 + x) + (8 + x)] / 2 = 14 + x.
5. Using Heron's formula:
   Area(ΔABC) = √[s(s − a)(s − b)(s − c)]
   Area(ΔABC) = √[(14 + x)(x)(8)(6)] = √[48x(14 + x)].
6. Area can also be calculated by splitting into three triangles (ΔOBC, ΔOCA, ΔOAB):
   Area(ΔABC) = ½ × radius × (a + b + c) = ½ × 4 × (28 + 2x) = 4(14 + x).
7. Equating both area formulas:
   √[48x(14 + x)] = 4(14 + x)
   48x(14 + x) = 16(14 + x)²
   3x = 14 + x ⇒ 2x = 14 ⇒ x = 7 cm.
8. Therefore:
   AB = 8 + 7 = 15 cm
   AC = 6 + 7 = 13 cm.

How to Prepare for Class 10 Circles in Board Exam 2025–2026

To secure full marks in the geometry section of CBSE Class 10 Mathematics, follow these tested preparation strategies:

1. Draw Clean, Labelled Diagrams: Every geometry solution must begin with a neat pencil sketch indicating the centre, radii, points of contact, and external points. In CBSE evaluation, marks are deducted if the geometric reference diagram is missing.
2. State Geometric Reasons in Brackets: When writing equality steps (e.g., AP = AS or ∠OPT = 90°), always state the theorem explicitly as a reason: [Tangents drawn from an external point are equal] or [Radius ⊥ Tangent at point of contact].
3. Master Algebraic Proofs: Questions involving circumscribed quadrilaterals and triangles frequently combine algebraic substitution with circle theorems. Practice writing each derivation in sequential equation format.

Where to Practice More

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