Let tangents at points A (4,-11) and B (8,-5) on circle x^2+y^2-3 x+10 y-15=0 , intersect at point C

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10-1.Circle and System of Circles

hard

Let the tangents at the points $A (4,-11)$ and $B (8,-5)$ on the circle $x^2+y^2-3 x+10 y-15=0$, intersect at the point $C$. Then the radius of the circle, whose centre is $C$ and the line joining $A$ and $B$ is its tangent, is equal to

A

$\frac{3 \sqrt{3}}{4}$

B

$2 \sqrt{13}$

C

$\sqrt{13}$

D

$\frac{2 \sqrt{13}}{3}$

(JEE MAIN-2023)

Solution

Equation of tangent at $A (4,-11)$ on circle is

$\Rightarrow 4 x-11 y -3\left(\frac{ x +4}{2}\right)+10\left(\frac{ y -11}{2}\right)-15=0$

$\Rightarrow 5 x -12 y -152=0 \ldots \ldots(1)$

Equation of tangent at $B (8,-5)$ on circle is

$\Rightarrow 8 x-5 y-3\left(\frac{x+8}{2}\right)+10\left(\frac{y-5}{2}\right)-15=0$

$\Rightarrow 13 x-104=0 \Rightarrow x=8$

put in $(1)$ $\Rightarrow y =\frac{28}{3}$

$r =\left|\frac{3.8+\frac{2.28}{3}-34}{\sqrt{13}}\right|=\frac{2 \sqrt{13}}{3}$

Standard 11

Mathematics

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