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Let PQ and RS be tangents at the extremi...

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

A

`sqrt(PQ.RS)`

B

`(PQ+RS)/(2)`

C

`(2PQ.RS)/(PQ+RS)`

D

`sqrt(PQ^(2)+RS^(2))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
From figure, it is clear that `Delta PRQ and Delta RSP` and similar.

`therefore(PR)/(RS)=(PQ)/(RP)`
`rArrPR^(2)=PQ*RS`
`rArrPR=sqrt(PQ*RS)`
`rArr 2r=sqrt(PQ*RS)`
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