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?

Let $(x_0,y_0)$ be the point of contact to $(x-1)^2+y^2=\frac85$. The tangent line equation is:
$$y=y_0+y'(x_0)(x-x_0) \Rightarrow \\
y=y_0+\frac{1-x_0}{y_0}(x-x_0) \Rightarrow \\
\frac{1-x_0}{y_0}x-y+y_0-\frac{1-x_0}{y_0}x_0=0 \Rightarrow \\
a=\frac{1-x_0}{y_0};b=-1;c=y_0-\frac{1-x_0}{y_0}x_0$$
The tangent line passes through the point $(-1,2)$ (the center of the circle $(x+1)^2+(y-2)^2=4$). 
So we make up the system:
$$\begin{cases}2=y_0+\frac{1-x_0}{y_0}(-1-x_0)\\ (x_0-1)^2+y_0^2=\frac85\end{cases}\Rightarrow (x_0,y_0)=(-\frac15,-\frac25),(\frac75,\frac65).$$
Hence:
$$(a,b,c)=(-3,-1,-1); (-\frac13,-1,\frac53).$$ 
