A very nice example is given by the Villarceau circles: a revolution torus is cut by a bitangent hyperplane along the union of two circles. You can of course make the computation, but when you know some projective algebraic geometry you can prove it in a few words. Roughly:

1. The revolution torus has an algebraic equation of degree four, so that it intersects any plane along a degree four curve.

2. If the plane is bitangent, then this curve has two double points so that it must be the union of two ellipses.

3. It is easily checked that in the complex world, the torus as well as the plane contain the circular points at infinity, so that in fact the ellipses are circles.