Q1. [1]
In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90°$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
- A $3\sqrt{2}$ cm
- B $6\sqrt{2}$ cm
- C 3 cm
- D 6 cm
Previously asked in CBSE board exam
2025 30/6/1 Q9
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Option (D) 6 cm
Since ∠P = 90° and PA = PB (tangent lengths), OAPB is a square. So OA = PA = OB = PB = radius r.
By Pythagoras in △PAB: AB² = PA² + PB² = r² + r² = 2r²
∴ $3\sqrt{2} = r\sqrt{2}$ ⟹ r = 3 cm. Diameter = 6 cm.
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Explanation
- Since ∠OAP = ∠OBP = 90° (radius ⊥ tangent) and ∠P = 90°, quadrilateral OAPB is a rectangle; further PA = PB implies it is a square, so PA = PB = OA = OB = r.
- Triangle PAB is right-angled at P, giving AB = r√2. Setting r√2 = 3√2 gives r = 3, diameter = 6 cm.
- Examiners expect you to identify OAPB as a square and apply Pythagoras to find the radius quickly.
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