Q1. [2] § 10.1 Introduction § 10.2 Tangent to a Circle § 10.4 Summary
A person is standing at $P$ outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points $A$ and $B$ on the ground are 10 m ($PA$ and $PB$ are tangents to the circle). Find the radius of the circular ground.
Previously asked in CBSE board exam
2025 30/1/1 Q25
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding rag
Model Answer
Given: Distance from external point P to centre O = 26 m; length of tangent PA = 10 m.
Since the tangent is perpendicular to the radius at the point of contact (Theorem 10.1):
$$OA^2 = OP^2 - PA^2$$
$$OA^2 = 26^2 - 10^2 = 676 - 100 = 576$$
$$OA = 24 \text{ m}$$
∴ The radius of the circular ground is 24 m.
Source: Chapter 10, Section 10.2 (Theorem 10.1)
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Explanation
- The key property used: tangent ⊥ radius at point of contact, forming a right angle at A.
- This makes triangle OAP a right triangle with hypotenuse OP = 26 m.
- Apply Pythagoras theorem directly. Examiners expect the formula written clearly, substitution shown, and the final answer stated.
- PA = PB (equal tangents), but only one tangent length is needed here.
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