Theorem. The intersection curve of a sphere and a plane is a circle.

Proof. We prove the theorem without the equation of the sphere. Let $c$ be the intersection curve, $r$ the radius of the sphere and $OQ$ be the distance of the centre $O$ of the sphere and the plane. If $P$ is an arbitrary point of $c$ , then $OPQ$ is a right triangle. By the Pythagorean theorem, $$PQ = \varrho = \sqrt{r^2\!-\!OQ^2} = \mbox{\;constant}.$$

Thus any point of the curve $c$ is in the plane at a constant distance $\varrho$ from the point $Q$ , whence $c$ is a circle.

Remark. There are two special cases of the intersection of a sphere and a plane: the empty set of points ($OQ > r$ ) and a single point ($OQ = r$ ); these of course are not curves. In the former case one usually says that the sphere does not intersect the plane, in the latter one sometimes calls the common point a zero circle (it can be thought a circle with radius 0).