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(C) Let $O$ be the center of the circle. $OR$ is the angle bisector of $\angle PRQ$. Let $M$ be the intersection of $PQ$ and $OR$. Since $RP=RQ$,$	ri

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$\frac{15}{4}$

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(C) Let $O$ be the center of the circle. $OR$ is the angle bisector of $\angle PRQ$. Let $M$ be the intersection of $PQ$ and $OR$. Since $RP=RQ$,$	riangle RPQ$ is an isosceles triangle,and $RM \perp PQ$ and $PM = MQ = \frac{1}{2} PQ = 3$.In right-angled $	riangle RPM$,by Pythagoras theorem:$RM^2 = PR^2 - PM^2 = 5^2 - 3^2 = 25 - 9 = 16 \Rightarrow RM = 4$.Let $\angle PRM = 	heta$. Then $	an 	heta = \frac{PM}{RM} = \frac{3}{4}$.In right-angled $	riangle OPR$ (where $\angle OPR = 90^\circ$ because $PR$ is a tangent),$\angle ORP = 	heta$. Thus,$	an 	heta = \frac{OP}{PR} = \frac{r}{5}$.Equating the two expressions for $	an 	heta$:$\frac{r}{5} = \frac{3}{4} \Rightarrow r = \frac{15}{4}$.

Tangents to a circle at points $P$ and $Q$ on the circle intersect at a point $R$. If $PQ=6$ and $PR=5$,then the radius of the circle is

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