Let PQ and RS be the tangents at the extremit | vedclass

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(C) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.In $	riangle PXR$ and

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$(PX)^2 + (RX)^2$

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(C) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.In $	riangle PXR$ and $	riangle QXP$,since $X$ lies on the circle and $PR$ is the diameter,$\angle PXR = 90^{\circ}$.By properties of similar triangles,$	riangle PQR \sim 	riangle RSP$ is not directly applicable,but we can use the property of tangents and the right angle at $X$.Since $\angle PXR = 90^{\circ}$,in $	riangle PXR$,by Pythagoras theorem,$(PR)^2 = (PX)^2 + (RX)^2$.From the geometry of the tangents,$	riangle P X Q$ and $	riangle R X S$ are related such that $PQ \cdot RS = (PR)^2$.Therefore,$PQ \cdot RS = (PX)^2 + (RX)^2$.

Let $PQ$ and $RS$ be the 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 $(PQ \cdot RS)$ is equal to

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