If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)? I have a question that goes: 
If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?
So what I tried was I know that since the line is normal to the 2nd circle, so it must pass through the center of the second circle which is $(-1,2)$.
So from that I got that $$-a+2b+c=0$$
But I cant really find any other equations here that would help, I tried differentiation the curves but I dont have the point of contact so can't really do anything there.
I also know that the tangent to the circle $x^2+y^2+2gx+2fy+c = 0$ at $(a,b)$ is $ax+by+(a+x)g+(b+y)f +c = 0$
I don't know how to proceed, can someone help?

By sketching a graph in mind, we see that the line $ax+by+c=0$ is just a line that is passing through $(-1,2)$ (the center of the circle $x^2+y^2+2x-4y+1=0$) and is tangent to the circle $x^2+y^2-2x=3/5$.
Therefore we looking for tangent lines from point $(-1,2)$ to the circle $x^2+y^2-2x=3/5$.
The line passing through $(-1,2)$ with slop $m$ is
$$y=m(x+1)+2.$$
The intersects points of the line and the circle $x^2+y^2-2x=3/5$ is found by
$$x^2+(m(x+1)+2)^2-2x=3/5$$
simplifying:
$$(1+m^2)x^2+[2m(m+2)-2]x+(m+2)^2-3/5=0\tag{*}$$
Since we need the line $y=m(x+1)+2$ to touch the circle, then the discriminant of $(*)$ must be zero:
$$(2m(m+2)-2)^2-4(1+m^2)((m+2)^2-3/5)=0$$
which simplifies to
$$\frac{-4}{5}(m+3)(m+\frac{1}{3})=0$$
i.e. $m=-3$ or $m=-1/3$. Thus the tangent lines are
$$y=-3(x+1)+2\quad\text{and}\quad y=-\frac13(x+1)+2$$
