I've tried to prove that there exists only one point $P$ on $O_1O_2$ such that $Pow(P,O_1)=Pow (P,O_2)$ where $O_1 $ and $O_2$ are circles with no point of tangency and I've got the following contradiction:
(Let $R$ and $r$ be the radii of circles $O_1$, and $O_2$ respectively)
$Pow(P,O_1)=Pow(P,O_2)$
$(PO_1)^2- R^2=(PO_2)^2 -r^2$
Then using the fact that $PO_1+PO_2=O_1O_2$ for $PO_1$ , I have:
$(O_1O_2)^2-2PO_2 \cdot O_1O_2 = R^2-r^2$
$O_1O_2(PO_1-PO_2)=R^2-r^2$
Now if I let $PO_1=R+x$ and $PO_2=r+y$ and consequently $O_1O_2=R+r+x+y$ I have
$[(R+x)+(r+y)] \cdot [R+x -(r+y)]=R^2-r^2$
and finally $(R+x)^2-(r+y)^2=R^2-r^2$ which must have solutions $x,y=0$ .
But doesn't that imply $P$ being the point of intersection of circles $O_1,O_2$?What am i doing wrong ?
I think that we cannot say that $$(R+x)^2-(r+y)^2=R^2-r^2\Rightarrow x=y=0.$$
Let $O_1O_2=s,O_1P=X$. Then, since $PO_2=s-X$, we have $$\begin{align}&O_1O_2(O_1O_2-2PO_2)=R^2-r^2\\&\Rightarrow s(s-2(s-X))=R^2-r^2\\&\Rightarrow s^2-2s(s-X)=R^2-r^2\\&\Rightarrow 2sX=R^2-r^2+s^2\\&\Rightarrow X=\frac{R^2-r^2+s^2}{2s}\end{align}$$ So, there exists only one such point.