<s>[INST] Proof: Two non identical circles have at most 2 same points I'm struggeling with an analytic proof for the fact, that two different circles have at most 2 same points. (I try to solve it analytical, because geometrical I already prooved it).
I tried to start with the equations
$r_1^2=(x-a_1)^2+(y-b_1)^2$ and $r_2^2=(x-a_2)^2+(y-b_2)^2$, further
$r_1^2=x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2$ and the same for the 2nd equation.
Then I get
\begin{eqnarray*}
r_1^2&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2\\
r_2^2&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2\\
0&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2\\
0&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Rightarrow& x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2=x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Leftrightarrow& 0=-2xa_1+a_1^2-2xb_1+b_1^2-r_1^2+2xa_2-a_2^2+2yb_2-b_2^2+r_2^2
\end{eqnarray*}
But now I don't know how to move on. Can someone give a hint?
 [/INST] You have worked too hard. The first circle has equation of the shape $x^2+y^2+ax+by+c=0$. The second circle has equation of the shape $x^2+y^2+dx+ey+f=0$. The two equations are not identical, since the circles are not.
If $(x,y)$ lies on both circles, then both equations are satisfied, and therefore (subtract) we have
$$(a-d)x+(b-e)y+(c-f)=0.\tag{$1$}$$
If $b-e=0$, then $x$ is determined, and substituting in the equation of one of the circles, we get a quadratic equation in $y$, which has at most $2$ solutions.
If $b-e\ne 0$, then we can use the linear equation to solve for $y$ in terms of $x$. Substitute in the equation of one of the circles. We get a quadratic equation in $x$, which has at most $2$ solutions. And from $(1)$, each corresponds to a unique value for $y$.
Note that we do not need to carry out the substitutions, all that is needed is to imagine  carrying them out.
Remark: The line with Equation $(1)$ is called the radical axis of the two circles. If the two circles do indeed meet in $2$ distinct points, the radical axis is the line through these $2$ points.    
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