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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 10 — Circles: the tangent–radius perpendicularity, the equal-tangents theorem, and tangent-length problems. Every problem uses a right angle between the tangent and the radius.
Key Concepts (Quick Recap)
- A tangent touches a circle at exactly one point.
- Tangent ⊥ radius at the point of contact.
- Length of tangent from an external point: L = √(d² − r²), where d = distance from centre, r = radius.
- Two tangents from an external point are equal in length.
Exercise 10.2 — Tangents
Q1. The length of a tangent from a point Q at distance 25 cm from the centre of a circle is 24 cm. Find the radius. By Pythagoras (tangent ⊥ radius): r² = 25² − 24² = 625 − 576 = 49 → r = 7 cm.
Q2. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at Q so that OQ = 12 cm. Find the length of PQ. PQ = √(OQ² − OP²) = √(12² − 5²) = √(144 − 25) = √119 = √119 cm ≈ 10.9 cm.
Q3. Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that touches the smaller circle. The chord is tangent to the smaller circle, so the radius (3 cm) to the point of contact is perpendicular to the chord and bisects it. Half-chord = √(5² − 3²) = √16 = 4 → chord = 2 × 4 = 8 cm.
Q4. From an external point, two tangents are drawn to a circle of radius 8 cm. If each tangent is 15 cm long, find the distance of the external point from the centre. d = √(r² + L²) = √(8² + 15²) = √(64 + 225) = √289 = 17 cm.
Theorem-Based Questions
Q5. Prove that the lengths of tangents drawn from an external point to a circle are equal. Let PA and PB be tangents from external point P to a circle with centre O. In △OAP and △OBP: OA = OB (radii); ∠OAP = ∠OBP = 90° (tangent ⊥ radius); OP = OP (common). By RHS congruence, △OAP ≅ △OBP, so PA = PB. ∎
Q6. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. Let the circle touch AB, BC, CD, DA at P, Q, R, S. Tangents from each vertex are equal: AP = AS, BP = BQ, CR = CQ, DR = DS. AB + CD = (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS). AD + BC = (AS + DS) + (BQ + CQ). Both sums equal AS + BQ + CQ + DS, so AB + CD = AD + BC. ∎
Q7. The tangent at a point C of a circle and a diameter AB, when extended, intersect at P. If ∠PCA = 110°, find ∠CBA. Join OC. OC ⊥ PC, so ∠OCP = 90°. ∠OCA = ∠PCA − ∠OCP = 110° − 90° = 20°. OA = OC (radii) → ∠OAC = ∠OCA = 20°. ∠CBA is the angle in the alternate segment; using the exterior-angle/isosceles relations, ∠CBA = 70° (since ∠ACB = 90° in the semicircle and ∠CAB = 20° give ∠CBA = 180° − 90° − 20° = 70°).
Common Mistakes
- Forgetting the 90° between tangent and radius — it is the key to using Pythagoras.
- Mixing up d (centre-to-point) and the tangent length L in L = √(d² − r²).
- In circumscribed-quadrilateral proofs, not pairing the equal tangents from each vertex correctly.
Exam Tips
- Draw the radius to the point of contact immediately — it creates the right triangle.
- "Tangents from an external point are equal" is used in almost every proof — state it explicitly.
- For a chord touching an inner concentric circle, the inner radius bisects the chord at right angles.