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# 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,

$PQ=\varrho=\sqrt{r^{2}\!-\!OQ^{2}}=\mbox{\;constant}.$ |

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).

## Mathematics Subject Classification

51M05*no label found*

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