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Question: Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius r. Such that PS ...

Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius r. Such that PS and RQ intersect at a point X on the circumference of the circle, then 2r equals –

A

PQ.PS\sqrt{PQ.PS}

B

PQ+RS2\frac{PQ + RS}{2}

C

2PQ.RSPQ+RS\frac{2PQ.RS}{PQ + RS}

D

PQ2+RS22\sqrt{\frac{PQ^{2} + RS^{2}}{2}}

Answer

PQ.PS\sqrt{PQ.PS}

Explanation

Solution

From the fig. we have

PQPR\frac{PQ}{PR} = tan (p/2 – q) = cot q

and RSPR\frac{RS}{PR} = tan q

Ž PQPR.RSPR\frac{PQ}{PR}.\frac{RS}{PR} = 1

Ž (PR)2 = PQ . PS

Ž (2r)2 = PQ . PS

Ž 2r = PQ.PS\sqrt{PQ.PS}.