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?

A circle is defined as the locus of a degree 2 polynomial.
The intersection of two circles is then, the roots of a degree 2 polynomial
There are at most $n$ roots of a degree $n$ polynomial
