An interesting problem on circles

Geometry Level pending

$PR$ is the diameter of a circle, and $PQ$ and $RS$ are tangents on the circle such that $Q$ and $S$ lie on the same side of the diameter $PR$ , and $QR$ and $PS$ intersect on the circumference of the circle. If $QP = 7$ , $SR = 9$ , find $PR^2$ .

The answer is 63.

1 solution

Thanos Petropoulos
Apr 5, 2021

Let $M$ be the intersection point of $QR$ and $PS$ . $M$ subtends diameter $PR$ by a right angle. Since, $PQ$ and $RS$ are tangent to the circle, $\angle QPR = \angle PRS = 90{}^\circ$

Hence,
$\angle RQP=90{}^\circ -\angle QPM=\angle SPR$

These imply that $\triangle RQP$ and $\triangle SPR$ are similar right triangles, thus $\frac{QP}{PR}=\frac{PR}{RS}\Rightarrow P{{R}^{2}}=QP\cdot RS\Rightarrow P{{R}^{2}}=7\times 9=\boxed{63}$

An interesting problem on circles

The answer is 63.

1 solution

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