Question

Let PQ and RS be tangents at the extremities of 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. $\frac{\sqrt{P{Q}^2+R{S}^2}}{2}$

Answer 1

The correct option is A $\sqrt{PQ.RS}$

In triangles PQR and PRS

$\angle PQR=\theta =\angle SPR$

$\angle QPR=\angle SRP={90}^{\circ }$

$⟹△PQR$ and $\angle RPS$ are similar.

$⟹\frac{PQ}{PR}=\frac{RP}{RS}$

$PR=\sqrt{PQ.RS}=2r$