Q. If a circle $C$ passing through $(4, 0)$ touches the circle $x^2 + y2 + 4x - 6y - 12 = 0$ externally at a point $(1, -1)$, then the radius of the circle $C$ is :
Solution:
Let $A$ be the centre of given circle and $B$ be the centre of circle $C$.
$x^2+y^2+4x-6y-12=0$
$\therefore A=\left(_{-2, 3}\right)$ and $B=\left(g. f\right)$
Now, from the figure, we have
$\frac{-2+g}{2}=1$ and $\frac{3+f}{2}=-1$
(By mid point formula)
$\Rightarrow g=4 $ and $f= - 5$
Now, required radius
$=OB=\sqrt{9+16}=\sqrt{25}=5$
