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 $RS \perp PR$.In $	riangle PXR$,$\a

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$\sqrt{PQ \cdot RS}$

Vedclass · Answer

(A) Let the diameter be $PR = 2r$. Since $PQ$ and $RS$ are tangents at $P$ and $R$ respectively,$PQ \perp PR$ and $RS \perp PR$.In $	riangle PXR$,$\angle PXR = 90^{\circ}$ because it is an angle in a semicircle.Let $\angle RPX = 	heta$. Then $\angle PRX = 90^{\circ} - 	heta$.In $	riangle PQR$,$\angle PQR = 90^{\circ} - 	heta$ and $\angle PRQ = 90^{\circ}$.Thus,$	an(\angle RPX) = 	an 	heta = \frac{RS}{PR} = \frac{RS}{2r}$.Also,in $	riangle PQR$,$	an(\angle PRQ) = 	an(90^{\circ} - 	heta) = \cot 	heta = \frac{PQ}{PR} = \frac{PQ}{2r}$.Therefore,$	an 	heta = \frac{2r}{PQ}$.Equating the two expressions for $	an 	heta$:$\frac{RS}{2r} = \frac{2r}{PQ} \implies (2r)^2 = PQ \cdot RS \implies 2r = \sqrt{PQ \cdot RS}$.

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

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