Question

If the locus of centre of circle which cuts the circles $x^2+y^2+4x-6y+9=0$ and $x^2+y^2-4x+6y+4=0$ orthogonally is $ax+by+c=0$ then $a+b+c$ is equal to

Answer 1

$\because$ Centre of circle which cuts the given two circles orthogonally lies on radical axis,
$\therefore$ Locus of centre $(h,k)$ will lies on $S_1-S_2=0$
$\Rightarrow 8h-12k+5=0$
So, locus is $8x-12y+5=0$
$\therefore a+b+c=8-12+5=1$