### Circle Question 4

#### Question 4 - 29 January - Shift 1

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

(1) $\frac{3 \sqrt{3}}{4}$

(2) $2 \sqrt{13}$

(3) $\sqrt{13}$

(4) $\frac{2 \sqrt{13}}{3}$

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Equation of Tangent in Point form, Distance between point and line

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

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

$\Rightarrow 5 x-12 y-152=0 \quad ….. (1)$

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

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

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

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

the line joining $A$ and $B$ is given by

$(y+11) = \frac{(-5+11)}{(8-4)}(x-4)$

$3x-2y-34=0$

Now, Distance between point $(8, \frac{-28}{3})$ and line $3x-2y-34=0$ is given by

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