Asked by Dillion

Let A and B be two points on the hyperbola xy=1, and let C be the reflection of B through the origin.

Let Gamma be the circumcircle of triangle ABC and let A' be the point on Gamma diametrically opposite A. Show that A' is also on the hyperbola xy=1.
Answered by Dillion
Nvrm I solved it.
Answered by oobleck
the hyperbola xy=1 has two branches, symmetric about the origin, since (-x)(-y) = 1
So, if A = (x,y), then A'=(-x,-y) and is also on the hyperbola.
There is no need to introduce any circle, since A' is directly opposite A.