I'm pretty sure I've got the first part of this proof correct, but I always seem to have problems proving the converse and this is no exception.


Here's the problem: Let S(r,P) denote the circle of radius r centred at P. Show that the reflection through the origin is again a circle.

Here's what I've done so far:

Let X be a point on the circle S(r,P). Then Norm(X-P) = r. Reflecting through the origin, O, we have X -> -X and P -> -P. Thus we have Norm(-X-(-P)) = Norm(P-X) = Norm(X-P) = r. Thus we see that -X is at a distance r from -P, and hence lies on a circle centred at -P with radius r, that is, on S(r,-P).


Conversely, given a point Y on the circle S(r,-P) we have, Norm(Y-(-P)) = Norm(Y+P) = r. Let X=-Y, then Y=-X is the reflection of X through O. Thus we have Norm(P-X) = r = Norm(X-P) and therefore every point on the circle S(r,-P) is the image under reflection through O of a point on the circle S(r,P).

Any general suggestions on proving the converse would be much appreciated.
Thanks,
Ultros