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<record version="19" id="10934">
 <title>intersection of sphere and plane</title>
 <name>IntersectionOfSphereAndPlane</name>
 <created>2008-08-11 11:49:58</created>
 <modified>2008-08-15 15:31:53</modified>
 <type>Theorem</type>
<parent id="186">sphere</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <author id="3771" name="CWoo"/>
 <classification>
	<category scheme="msc" code="51M05"/>
 </classification>
 <defines>
	<concept>zero circle</concept>
 </defines>
 <related>
	<object name="ConformalityOfStereographicProjection"/>
 </related>
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\newtheorem{thm}{Theorem}
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\newtheorem{cor}{Corollary}</preamble>
 <content>\textbf{Theorem.}\, The intersection curve of a sphere and a plane is a circle.

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

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\rput(0.3,0){$O$}
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\rput(-2.45,0.68){$P$}
\rput(0.3,1.23){$Q$}
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Thus any point of the curve $c$ is in the plane at a \PMlinkescapetext{constant} distance $\varrho$ from the point $Q$, whence $c$ is a circle.\\

\textbf{Remark.}\, There are two special cases of the intersection of a sphere and a plane:\, the empty set of points ($OQ &gt; r$) 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 {\em zero circle} (it can be thought a circle with radius 0).</content>
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