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Question

Mathematics Question on Conic sections

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

PQ,RS\sqrt{PQ, RS}

B

PQ+RS2\frac {PQ+RS}{2}

C

2PQ−RSPQ+RS\frac {2PQ-RS}{PQ+RS}

D

PQ2+RS22\sqrt{\frac {{PQ}^2+{RS}^2}{2}}

Answer

PQ,RS\sqrt{PQ, RS}

Explanation

Solution

From figure, it is clear that △\triangle PRQ and △\triangle RSP are
similar,
∴\therefore PRRS=PQRP⇒PR2=PQ.RS\frac {PR}{RS}=\frac {PQ}{RP}\Rightarrow {PR}^2=PQ . RS
⇒PR=PQ.RS\Rightarrow PR=\sqrt{PQ . RS}
⇒2r=PQ.RS\Rightarrow 2r=\sqrt{PQ . RS}