Q1. [1] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.4 Summary
In the adjoining figure, AB is the chord of the larger circle touching the smaller circle. The centre of both the circles is O. If AB = $2r$ and OP = $r$, then the radius of larger circle is :
- (a) $2r$
- (b) $3r$
- (c) $2\sqrt{2}\,r$
- (d) $\sqrt{2}\,r$
Previously asked in CBSE board exam
2025 30/4/1 Q14
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
(d) $\sqrt{2}\,r$
Since AB is tangent to the smaller circle at P, OP ⊥ AB (radius ⊥ tangent). AB = 2r, so PB = r. In right △OPB: $OB^2 = OP^2 + PB^2 = r^2 + r^2 = 2r^2$, giving $OB = r\sqrt{2}$.
Explanation
Key steps: tangent ⊥ radius (Theorem 10.1) makes △OPB right-angled at P; perpendicular from centre bisects chord, so PB = r; then Pythagoras gives the radius of the larger circle as $\sqrt{2}\,r$. Examiners expect the right-angle reasoning and Pythagoras application clearly stated.
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