Q.

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

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a

PQ+RS2

b

PQ2+RS22

c

PQ.RS

d

2PQ.RSPQ+RS

answer is A.

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Detailed Solution

tanθ=PQPR=PQ2r also tanπ2θ=RS2r
cotθ=RS2rtanθcotθ=PQRS4r24r2=PQRS2r=PQRS

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