# 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).

Title | intersection of sphere and plane |
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Canonical name | IntersectionOfSphereAndPlane |

Date of creation | 2013-03-22 18:18:39 |

Last modified on | 2013-03-22 18:18:39 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 22 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M05 |

Related topic | ConformalityOfStereographicProjection |

Defines | zero circle |