We have circle AB, an arbitrary point D on circle AB, and a free point C.

E is the reflection point of C with respect to the circle AB. That is, AC*AE=(r of circle AB)^2.

Circle OD passes through point C, and point D, and is perpendicular to circle AB.

Two circles perpendicular to each other means their repective radius connecting the center and the intersection point are perpendicular.

By moving point D around, we can see circle OD always goes through point E too.

We can prove:

All circles through point C and perpendicular to circle AB also goes through C's reflection point E.

More information can be found at http://www.cut-the-knot.org/Curriculum/Geometry/SymmetryInCircle.shtml