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

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a

PQ.RS

b

PQ+RS2

c

2PQ . RSPQ+RS

d

PQ2+RS22

answer is A.

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Detailed Solution

∠PXR=90° So, ∠XPR=θ⇒∠XRP=90°−θ∴tanθ=SRPR=SR2r--------(1)and tan90°-θ=PQPR=PQ2r⇒cotθ=PQ2r---------2⇒1=PQRS(2r)2 by 1 and 2⇒2r=PQ·RS

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