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Circles
EXERCISE 10.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm (B) 13 cm (C) 8.5 cm (D) 119 cm.

Step-by-Step Solution

Key Concept: The radius drawn to the point of tangency is perpendicular to the tangent. Hence, triangle OPQ is a right‑angled triangle with right angle at P. Use the Pythagorean theorem to relate the sides OP (radius), OQ (given) and PQ (required).
1. Draw the circle with centre O and radius OP = 5 cm.
2. At point P draw the tangent PQ.
3. Join O to Q; the line OQ passes through the centre and meets the tangent at Q.
4. Since a radius to a point of tangency is perpendicular to the tangent, \(OP \perp PQ\).
5. Therefore, triangle \(OPQ\) is right‑angled at \(P\).
6. Apply the Pythagorean theorem:
$$OP^{2} + PQ^{2} = OQ^{2}$$
Substituting the known lengths:
$$5^{2} + PQ^{2} = 12^{2}$$
$$25 + PQ^{2} = 144$$
$$PQ^{2} = 144 - 25 = 119$$
7. Hence,
$$PQ = \sqrt{119}\ \text{cm}$$
8. The numerical value of \(\sqrt{119}\) is approximately 10.9 cm, which is not listed among the options; the closest representation is option (D) "119 cm" interpreted as \(\sqrt{119}\) cm.

Correct Answer: $\sqrt{119}\text{ cm}$
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