I'm having a hard time proving the following:
Two circles with centers $P_1$ and $P_2$ respectively have two intersections $A$ and $B$. Prove that the line through $P_1$ and $P_2$ and the line through $A$ and $B$ are perpendicular.
I know that this has to be fairly simple, but every approach I see is somewhat circular. How to prove it rigorously?
1) $\triangle ABD$ is congruent with the $\triangle ACD$. (This is very obvious because this figure is symmetric about the line AD)
2) therefore $\angle BAE= \angle CAE$.
3) Hence $\triangle ABE$ and $\triangle ACE$ are congruent.
4) Hence $\angle BEA = \angle CEA$.
But $\angle BEA + \angle CEA =180^o$
Hint: show that $P_1AP_2=P_1BP_2$ and then prove that $P_1AH=P_1BH$, where $H$ is the intersection of lines $P_1P_2$ and $AB$.
Geometrical Approach: For ease of calculation, let the circle $C_1$, with radius $r_1$, be centered at the origin $(0, 0)$ & the circle $C_2$, with radius $r_2$, be centered on the x-axis at some point say $(a, 0)\ \forall\ \color{red}{0<a<(r_1+r_2)}$
Then we have following equations of the circles $$x^2+y^2=r_1^2\tag 1$$ $$(x-a)^2+y^2=r_2^2\tag 2$$ solving (1) & (2) we get $$(x-a)^2+r_1^2-x^2=r_2^2$$$$\implies x=\frac{a^2+r_1^2-r_2^2}{2a}=k\ (\text{constant})>0$$ Since, $r_1$, $r_2$ & $a$ are constants. substituting $x=k$ in (1), we get $$y^2=r_1^2-k^2\implies y=\pm\sqrt{r_1^2-k^2}$$
Now, the coordinates of the points of intersection $A$ & $B$ of circles are $(k, \sqrt{r_1^2-k^2})$ & $(k, -\sqrt{r_1^2-k^2})$ respectively then the slope of the line AB
$$=\frac{\sqrt{r_1^2-k^2}-\left(-\sqrt{r_1^2-k^2}\right)}{k-k}=\infty$$ Above result shows that the line AB is parallel to the y-axis i.e. line AB is perpendicular to the x-axis (i.e. line $P_1P_2$)
Hence, the line through intersection points $A$ & $B$ is perpendicular to the line $P_1P_2$ joining the centers of the circles.
