Question

If tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$ at point $(h,k),$ then the value of $h+k$ is

Answer 1

$S_1:x^2+y^2=5$
$S_2:x^2+y^2-8x+6y+20=0$

Equation of tangent at $(1,-2)$ to $S_1$ is
$x-2y-5=0$

Put $x=2y+5$ in $S_2,$ we get
$(2y+5{)}^2+y^2-8(2y+5)+6y+20=0$
$\Rightarrow 5y^2+10y+5=0$
$\Rightarrow (y+1{)}^2=0$
$\Rightarrow y=-1$
$\Rightarrow x=-2+5=3$
So, point of contact is $(3,-1).$
$\therefore h+k=2$