mathematics-class-11-unit-12-chapter-5-circles-l-5-13-cbvmloyrsvc

### <Success>Solution</Success>
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<p>$ C=\frac{-8 \sqrt{3} \pm \sqrt{192-\frac{320}{3}}}{8}$</p>
<p>i.e.,</p>
<p>$ C=-\sqrt{3} \pm \sqrt{3-\frac{5}{3}}$</p>
<p>$ =-\sqrt{3} \pm \frac{2}{\sqrt{3}}$</p>
<p>$ C=-\sqrt{3}+\frac{2}{\sqrt{3}},-\sqrt{3}-\frac{2}{\sqrt{3}}$</p>
<p>$ C=\frac{-1}{\sqrt{3}}, \frac{-5}{\sqrt{3}},\quad\quad  y=\frac{x}{\sqrt{3}}+c$</p>
<p>$ y=\frac{x}{\sqrt{3}}-\frac{1}{\sqrt{3}} \quad i.e., \quad x-\sqrt{3} y=1 , \quad x-\sqrt{3} y=5$</p>
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<h3 id="successexamplesuccess"><Success>Example</Success></h3>
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<p>2.) Let $P Q$ and $R S$ be tangents at the Extremities of the diameter $P R$ of a circle of radius $r$. If $P S$ and $R Q$ intersect-at a point $x$ on the circumference of the circle, then $2 r$ equals</p>
<p>a) $\sqrt{P Q \cdot RS}$</p>
<p>b) $\frac{P Q+R S}{2}$</p>
<p>c) $\frac{2 P Q \cdot R S}{P Q+R S}$</p>
<p>d) $\sqrt{\frac{P Q^2+R S^2}{2}}$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d191b5b2-b7dc-44b9-724e-e1a232b67b00/public"></p>

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### <Success>Example</Success>
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<p>$P T^2+O T^2=O P^2, \quad O T^2=4$</p>
<p>$ O P^2=(5-2)^2 +(3-(-8))^2 =34$</p>
<p>$ \frac{y-(-2)}{x-2}=\frac{5-(-3)}{9-2}$</p>
<p>$y=x-4$</p>
<p>$ (x-5)^2+(y-3)^2=4$</p>
<p>$ (x-5)^2+(x-7)^2=4$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d85e0a68-9d66-4bda-ade8-69da13b01900/public"></p>

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<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>$A(5,1)\quad B(7,3)$</p>
<p>$P T^2  =30$</p>
<p>$P A  =\sqrt{(1-(-2))^2+(5-2)^2} \\ \quad  =\sqrt{18}$</p>
<p>$P B  =\sqrt{(3-(-2))^2+(7-2)^2} \\ \quad =\sqrt{50}$</p>
<p>$P A .PB =\sqrt{18 \times 50} \\ \quad =30 $</p>
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