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Revision difference : intersection of sphere and plane |
| Version 5 |
Version 4 |
| \textbf{Theorem.}\, The intersection curve of a sphere and a plane is a circle. |
\textbf{Theorem.}\, The intersection curve of a sphere and a plane is a circle. |
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| {\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, |
{\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, |
| $$PQ = \varrho = \sqrt{r^2\!-\!OQ^2} = \mbox{\;constant}.$$ |
$$PQ = \varrho = \sqrt{r^2\!-\!OQ^2} = \mbox{\;constant}.$$ |
| 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. |
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. |
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