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Chapter 10: Circles

Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)

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NCERT Class 10 Maths | Chapter 10: Circles – Complete Solutions
Exercise 10.1 (Page 209)
11 mark
How many tangents can a circle have?
Answer: A circle can have infinitely many tangents.
21 mark each
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 ______.
Answer:
(i) one
(ii) secant
(iii) two
(iv) point of contact
32 marks
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. Find the length of PQ.
Answer:
In right triangle OPQ, OP = 5 cm (radius), OQ = 12 cm.
PQ = √(OQ² - OP²) = √(144 - 25) = √119 cm.
Length PQ = √119 cm.
42 marks
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.
Answer: [Diagram description] Draw a circle. Draw a line l outside the circle as given line. Draw a line m parallel to l that touches the circle at one point (tangent). Draw another line n parallel to l that cuts the circle at two points (secant).
Exercise 10.2 (Page 213)
11 mark
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. Find the radius of the circle.
Answer:
Let radius = r. Tangent PT = 24 cm, OQ = 25 cm.
In right triangle OTQ, r² = OQ² - PT² = 625 - 576 = 49 → r = 7 cm.
Radius = 7 cm.
21 mark
In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then find ∠PTQ.
Answer:
∠OPQ = ∠OQT = 90° (radius ⊥ tangent). In quadrilateral OPTQ, sum of angles = 360°.
90° + 90° + 110° + ∠PTQ = 360° → ∠PTQ = 70°.
∠PTQ = 70°.
31 mark
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 find ∠POA.
Answer:
∠APB = 80°. Since ∠OPA = ∠OPB = half of ∠APB? Actually, ∠APO = ∠BPO = 40°.
In right triangle OAP, ∠OAP = 90°, so ∠POA = 50°.
∠POA = 50°.
42 marks
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer:
Let AB be a diameter. Tangents at A and B are perpendicular to OA and OB respectively. Since OA and OB are in the same line, both tangents are perpendicular to the same line, hence they are parallel.
52 marks
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer:
The radius to the point of contact is perpendicular to the tangent. The perpendicular at the point of contact is unique; hence it must coincide with the radius, which passes through the centre.
63 marks
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.
Answer:
r² = OA² - AT² = 5² - 4² = 25 - 16 = 9 → r = 3 cm.
Radius = 3 cm.
73 marks
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.
Answer:
Let chord AB of larger circle touch smaller circle at C. OC = 3 cm, OA = 5 cm. AC = √(5² - 3²) = √16 = 4 cm.
Length of chord = 2 × AC = 8 cm.
Length = 8 cm.
83 marks
A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.
Answer:
Tangents from a point to a circle are equal.
From A: AP = AS; from B: BP = BQ; from C: CQ = CR; from D: DR = DS.
Adding: (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) → AB + CD = AD + BC.
93 marks
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°.
Answer:
Join OC. In triangles OPA and OCA, OP = OC (radii), OA common, ∠OPA = ∠OCA = 90° (tangents). So ΔOPA ≅ ΔOCA → ∠POA = ∠COA.
Similarly, ΔOQB ≅ ΔOCB → ∠QOB = ∠COB. Since XY || X'Y', ∠POA + ∠QOB = 90°? Actually, ∠POA + ∠COA + ∠COB + ∠QOB = 180°. But ∠POA = ∠COA and ∠COB = ∠QOB → 2∠COA + 2∠COB = 180° → ∠COA + ∠COB = 90° → ∠AOB = 90°.
103 marks
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.
Answer:
Let tangents from P touch circle at A and B. OA ⟂ PA, OB ⟂ PB. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. So ∠AOB + ∠APB = 180°. Hence they are supplementary.
112 marks
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:
Let parallelogram be ABCD. From equal tangents: AB + CD = AD + BC. But AB = CD and AD = BC (parallelogram). So 2AB = 2AD → AB = AD. Hence all sides equal → rhombus.
124 marks
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.
Answer:
Let tangents from B be BD = BF = 8 cm; from C, CD = CE = 6 cm; from A, AF = AE = x.
Sides: AB = x+8, AC = x+6, BC = 14.
Semi-perimeter s = (x+8 + x+6 + 14)/2 = (2x+28)/2 = x+14.
Area Δ = r·s = 4(x+14). Also by Heron: Area = √[s(s-a)(s-b)(s-c)].
Solve: x = 7. Then AB = 15 cm, AC = 13 cm.
Key Concepts Summary
Theorem 1Tangent Perpendicular to Radius
State the theorem relating a tangent and the radius at the point of contact.
Answer: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Theorem 2Lengths of Tangents
What can you say about the lengths of tangents drawn from an external point?
Answer: The lengths of the two tangents drawn from an external point to a circle are equal.

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