Question

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. $\frac {PQ+RS}{2}$
• C. $\frac {2PQ-RS}{PQ+RS}$
• D. $\sqrt{\frac {{PQ}^2+{RS}^2}{2}}$

Answer 1

The Correct Option is A

From figure, it is clear that $\triangle$ PRQ and $\triangle$ RSP are
similar,
$\therefore$ $\frac {PR}{RS}=\frac {PQ}{RP}\Rightarrow {PR}^2=PQ . RS$
$\Rightarrow PR=\sqrt{PQ . RS}$
$\Rightarrow 2r=\sqrt{PQ . RS}$