Inversion problem: Given 2 intersecting circles, orthogonal to a third, prove that the points of intersection and center of the third are collinear. The problem goes as follows:
Given two circles ($C_1$ and $C_2$) which intersect at $A$ and $B$ and are also orthogonal to a third cirlce $C_3$ with center $O$, prove that $A$, $B$ and $O$ are collinear.
Image of the geometric construction of the problem.
Here's what I have so far:
I thought inversion could help here as when I invert with A as my center of inversion, $C_1$ and $C_2$ become two straight lines ($C_1'$ and $C_2'$) which are orthogonal to the circle $C_3'$. $C_1'$ and $C_2'$ also pass through the image of $B$ after the inversion ($B'$). This implies that $B'$ is the center of $C_3'$. Image after the inversion.
This must somehow imply what I'm looking for, that $O$, $A$ and $B$ are collinear but I do not know how to conclude.
 A: Consider instead the inversion through the circle $C_3$.
Since $C_1$ intersects $C_3$ at two points $C,D$ in a right angle, the image of $C_1$ after the inversion is a circle that intersects $C_3$ at those same points and also in a right angle. So $C_1' = C_1$.
Similarly , $C_2' = C_2$. And this is enough to deduce that the inversion swaps $A$ with $B$, which implies that $A,B,O$ are colinear.
A: You have that $B'$ is the center of $C'_3$. But the center of $C_3$ and
the center of $C'_3$ are collinear with $A$,
therefore $A$, $B'$, and $O$ are collinear.
Also $B$ and $B'$ are collinear with $A$.
Therefore $B$ and $O$ are each on the line $AB'$, therefore $A$, $B$, and $O$
are collinear.

You could also use the Secant Theorem.
Let the line $OB$ intersect circle $C_1$ for a second time at $A_1$.
Then $(OB)(OA_1) = r^2$ where $r$ is the radius of $C_3$
(since the tangents from $O$ to $C_1$ touch $C_1$ at the intersection
with $C_3$).
Likewise if $OB$ intersect circle $C_2$ for a second time at $A_2$.
Then $(OB)(OA_2) = r^2$.
Therefore $OA_1= OA_2$, but both $A_1$ and $A_2$ lie on line $OB$ on the
same side of $O$, so $A_1$ and $A_2$ are the same point and this is
also the intersection point $A$.
