Theorem

Let P be a point on the radical axis of an elliptic family. Given any circle in the elliptic family, draw PT so that PT is tangent to the circle. The set of all such points T lie on a circle, C, with center P which is orthogonal to each circle in the elliptic family that C intercepts.

Diagram

Proof

PA = PB for all circles C1, C2, C3, and all others in the elliptical family.

PT1^2 = PT2^2 = PT3^2 = . . . since by a previous theorem

therefore PT1 = PT2 = PT3 . . .

Since all such points T(i) are equidistant from p, a circle O is formed with center P.

Circle O is then orthogonal to all circles in the elliptic family C1, C2, C3 . . . C(i) . . .

because the tangent T(i) passes through center P of circle O. (definition of orthogonal).

Q.E.D.