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.
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.