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

Options

A. `sqrt(PQ.RS)`
B. `(PQ+RS)/2`
C. `(2PQ.RS)/(PQ+RS)`
D. `sqrt((PQ^(2)+RS^(2))/2)`

Explanation

In triangles PQR and PRS, we have
`tan(pi/2-theta)=(PQ)/(PR)` and `tantheta=(RS)/(PR)`
`rArrcotthetaxxtantheta=(PQ)/(PR)xx(RS)/(PR)`
`rArrPR^(2)=PQ.RSrArrPR=sqrt(PQ.RS)rArr2r=sqrt(PQ.RS)`