Let PQ and RS be tangents at the extremities | vedclass

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(A) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.Let $\angle PRQ = 	he

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$\sqrt{PQ \cdot RS}$

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(A) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.Let $\angle PRQ = 	heta$. In $	riangle PQR$,$	an 	heta = \frac{PQ}{PR}$,so $PR = PQ \cot 	heta$.Since $X$ lies on the circle and $PR$ is the diameter,$\angle PXR = 90^{\circ}$.In $	riangle PXR$,$\angle XPR = 90^{\circ} - 	heta$ and $\angle XRP = 	heta$.In $	riangle PXS$,$\angle XPS = 90^{\circ} - 	heta$ and $\angle XSP = 	heta$. Thus,$	an 	heta = \frac{RS}{PR}$,so $PR = RS 	an 	heta$.Equating the two expressions for $PR$:$PQ \cot 	heta = RS 	an 	heta$$	an^2 	heta = \frac{PQ}{RS} \Rightarrow 	an 	heta = \sqrt{\frac{PQ}{RS}}$.Substituting this back into $PR = RS 	an 	heta$:$PR = RS \cdot \sqrt{\frac{PQ}{RS}} = \sqrt{PQ \cdot RS}$.Since $PR = 2r$,we have $2r = \sqrt{PQ \cdot RS}$.

Let $PQ$ and $RS$ be tangents at the extremities of a diameter $PR$ of a circle of radius $r$ such that $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle,then $2r$ equals

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