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?

Another way.
Let $y=mx+n$ be an equation of the tangent.
Thus, $$mx-y+n=0,$$ which since equations of our circles they are
$$(x+1)^2+(y-2)^2=4$$ and $$(x-1)^2+y^2=\frac{8}{5},$$ we obtain that the point $(-1,2)$ is placed on the line $y=mx+n$ 
and the distance from $(1,0)$ to the line is equal to $\sqrt{\frac{8}{5}}.$
Thus, 
$$-m-2+n=0$$ and
$$\frac{|m\cdot1-1\cdot0+n|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}},$$ which gives
$$\frac{|2m+2|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}}$$ or
$$5(m+1)^2=2(m^2+1).$$ 
Can you end it now?
