Let PQ and RS be tangents at the extremities | vedclass

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(A) Let $\angle PRQ = 	heta$. Since $PQ$ is a tangent at $P$ and $PR$ is the diameter,$\angle RPQ = 90^{\circ}$.In $	riangle RPQ$,$	an 	heta = \fr

Vedclass · Accepted Answer

$\sqrt{PQ \cdot RS}$

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(A) Let $\angle PRQ = 	heta$. Since $PQ$ is a tangent at $P$ and $PR$ is the diameter,$\angle RPQ = 90^{\circ}$.In $	riangle RPQ$,$	an 	heta = \frac{PQ}{PR} = \frac{PQ}{2r}$.Since $RS$ is a tangent at $R$ and $PR$ is the diameter,$\angle PRS = 90^{\circ}$.Given $X$ is on the circumference,$\angle RXP = 90^{\circ}$ (angle in a semicircle).In $	riangle RXP$,$\angle XRP = 90^{\circ} - 	heta$.In $	riangle PRS$,$	an(90^{\circ} - 	heta) = \frac{RS}{PR} = \frac{RS}{2r}$.Thus,$\cot 	heta = \frac{RS}{2r}$.Multiplying the two expressions: $	an 	heta \cdot \cot 	heta = \left(\frac{PQ}{2r}ight) \cdot \left(\frac{RS}{2r}ight) = 1$.$1 = \frac{PQ \cdot RS}{4r^2} \implies 4r^2 = PQ \cdot RS \implies 2r = \sqrt{PQ \cdot RS}$.

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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