We are given two circles $S_1$($x^2+y^2-2x-3=0$) and $S_2$($x^2+y^2-4y-6=0$). A line $ax+by=2$ which touches the former circle and is normal to the latter. We have to find the value of a and b.
I was able to find the value of b:-
the center of $S_2$ is (-g,-f)=(0,2) Now since the line is normal to this circle, it passes through the circle :- $$y-2=m(x-0)$$ $$y=mx+2$$ So the value of b is 1. I don't know how to find the value of m.