Question

Let the tangents at the points $A(4,-11)$ and $B(8,-5)$ on the circle $x^2+y^2-3x+10y-15=0$, 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. $2\sqrt{13}$
• B. $\frac{2\sqrt{13}}{3}$
• C. $\sqrt{13}$
• D. $\frac{3\sqrt{3}}{4}$

Answer 1

Answer: (B)

Equation of tangent at $A(4,-11)$ on circle is
$\Rightarrow 4x-11y-3\left(\frac{x+4}{2}\right)+10\left(\frac{y-11}{2}\right)-15=0$
$\Rightarrow 5x-12y-152=0\dots \dots (1)$
Equation of tangent at B $(8,-5)$ on circle is
$\Rightarrow 8x-5y-3\left(\frac{x+8}{2}\right)+10\left(\frac{y-5}{2}\right)-15=0$
$\Rightarrow 13x-104=0\Rightarrow x=8$
put in (1) $\Rightarrow y=\frac{28}{3}$
$r=\left|\frac{3.8+\frac{2.28}{3}-34}{\sqrt{13}}\right|=\frac{2\sqrt{13}}{3}$