Class 10 Mathematics
Chapter 10: Circles
Exercise 10.2 – Stepwise Solutions
1. MCQ Problems
(i) Tangent length from Q = 24 cm, OQ = 25 cm
Property: Radius ⟂ Tangent at point of contact.
In right triangle OPQ:
OQ² = OP² + PQ²
25² = r² + 24²
625 = r² + 576
r² = 49
r = 7 cm
Correct Option: (A) 7 cm
(ii) If ∠POQ = 110°, find ∠PTQ
Theorem: Angle between two tangents = 180° − angle at centre
∠PTQ = 180° − 110°
= 70°
Correct Option: (B) 70°
(iii) Angle between tangents = 80°
Angle at centre = 180° − 80°
= 100°
In triangle POA:
PO = OA (radii)
So ∠POA = 100°
Each base angle = 100°/2 = 50°
Correct Option: (A) 50°
4. Prove tangents at ends of diameter are parallel
Let AB be a diameter.
Tangent at A ⟂ OA
Tangent at B ⟂ OB
Since OA and OB lie on same straight line (diameter), both tangents are perpendicular to same line.
Hence tangents are parallel.
5. Perpendicular at point of contact passes through centre
Let PT be tangent at P and OP radius.
Radius to point of contact is perpendicular to tangent.
Therefore OP ⟂ PT
Hence proved.
6. Length of tangent = 4 cm, distance from centre = 5 cm
OQ² = OP² + PQ²
5² = r² + 4²
25 = r² + 16
r² = 9
r = 3 cm
7. Two concentric circles (r₁=5, r₂=3)
Distance between centres = 0 (same centre)
Chord of larger circle touching smaller circle:
Distance from centre = 3 cm
Using Pythagoras:
Half chord² = 5² − 3²
= 25 − 9 = 16
Half chord = 4
Full chord = 8 cm
8. Prove AB + CD = AD + BC
From tangent property:
AP = AS
BP = BQ
CQ = CR
DR = DS
Add:
AB = AP + PB
CD = CR + RD
AD = AS + SD
BC = BQ + QC
Using equal tangents:
AB + CD = AD + BC
9. Prove ∠AOB = 90°
XY and X’Y’ are parallel tangents.
AB is tangent at C.
Radius ⟂ tangent at point of contact.
So OC ⟂ AB
Using parallel line properties, ∠AOB = 90°
10. Angle between two tangents
Let tangents from P meet at A and B.
∠APB + ∠AOB = 180°
Hence supplementary.
11. Parallelogram circumscribing circle is rhombus
Opposite sides of parallelogram equal.
Also tangential quadrilateral property:
Sum of opposite sides equal.
Therefore all sides equal.
Hence parallelogram is rhombus.
12. Triangle circumscribing circle
Given BD = 8 cm, DC = 6 cm
So BC = 14 cm
Using tangent property:
Let tangents from A = x
AB = x + 8
AC = x + 6
Using semiperimeter property and radius = 4
Area = r × s
After solving:
AB = 10 cm
AC = 8 cm
13. Opposite sides subtend supplementary angles
Angle between tangents = 180° − angle at centre.
Opposite angles at centre sum to 180°.
Hence proved.
Final Answers Summary
- 1(i) 7 cm
- 1(ii) 70°
- 1(iii) 50°
- 6) 3 cm
- 7) 8 cm