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?
 A: 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
$$y=-3(x+1)+2\quad\text{and}\quad y=-\frac13(x+1)+2$$
A: 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:
$$(a,b,c)=(-3,-1,-1); (-\frac13,-1,\frac53).$$ 
A: 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?
