intersection of sphere and plane

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,

$$PQ=\varrho =\sqrt{r^2-O{Q}^2}=\text{ constant}.$$

Thus any point of the curve $c$ is in the plane at a 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).