Inversion with respect to a circle and orthogonal circles.

Theorem: If C2 is orthogonal to C1 (with center O) and A is a point inside C1 that is on C2 then the ray OA will intersect C2 at the point A' where A and A' are inverses with respect to the circle C1.

Proof: Consider the following figure.

Sorry, this page requires a Java-compatible web browser.

We need to show that m(OA)m(OA') = m(OP)².
Let m(OA)=a, m(AM)=m, m(OP)=R, m(O'M)=h, m(O'A) = S, m(OO') = T.
Then we want to show that a(a+2m) = R^2.

But a(a+2m) = a^2 + 2am.
And (a+m)^2+h^2 = T^2.
So a^2 + 2am +m^2 + h^2 = T^2.
But S^2 = m^2 + h^2, so
a^2 +2am + S^2 = T^2
a(a+2m) = T^2 - S^2 = R^2.

IRMC.