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

\frac{13}{3}

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4

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\frac{15}{4}

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\frac{16}{5}

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Solution

The correct option is D $\frac{15}{4}$

Given : $P Q = 6$ and $P R = 5$
So, $Q R = P R = 5$
Using Pythagoras theorem in $△ Q S R$
${Q S}^{2} + {S R}^{2} = {Q R}^{2} 9 + {S R}^{2} = 25 \Rightarrow S R = 4$

In $△ O S Q$
${O S}^{2} + {S Q}^{2} = {O Q}^{2} {O S}^{2} + 9 = r^{2}$
$x^{2} + 9 = r^{2}$ ..... $(i)$

In $△ O Q R$
${O Q}^{2} + {Q R}^{2} = {O R}^{2} r^{2} + 25 = {(O S + S R)}^{2} r^{2} + 25 = {(x + 4)}^{2} x^{2} + 9 + 25 = x^{2} + 16 + 8 x \Rightarrow x = \frac{9}{4}$
${(\frac{9}{4})}^{2} + 9 = r^{2}$ .... From $(i)$
$\Rightarrow r = \frac{15}{4}$
So option $C$ is correct

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

The correct option is D 154Given : PQ=6 and PR=5So, QR=PR=5Using Pythagoras theorem in △QSRQS2+SR2=QR29+SR2=25⇒SR=4In △OSQOS2+SQ2=OQ2OS2+9=r2 x2+9=r2 ..... (i)In △OQROQ2+QR2=OR2r2+25=(OS+SR)2r2+25=(x+4)2x2+9+25=x2+16+8x⇒x=94(94)2+9=r2 .... From (i)⇒r=154So option C is correct