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 Post subject: Equation of spherePosted: Wed Dec 16, 2015 11:52 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Given the circle with center ${\rm K} (5, 4, 0)$ and radius $r=6$ in the $xy$ plane , find the sphere that passes through that circle and touches the plane $\mathbb{P}: 3x+2y+6z=1$.

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 Post subject: Re: Equation of spherePosted: Wed Sep 14, 2016 8:26 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Let $\rm{K}_1$ be the center of the sphere $S$ in question. This must be on the line which is perpendicular to the $xy$-plane, where lies the given circle $C$ with center ${\rm K} (5, 4, 0)$ and radius $r=6$. So $\rm{K}_1(5,4,c)$. Let $R$ be the radius of the sphere $S$ and $\rm{M}$ any point of the circle $C$.
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sphere_thrght_circle.png [ 65.16 KiB | Viewed 520 times ] sphere_thrght_circle.png [ 65.16 KiB | Viewed 520 times ]

Then, by Pythagorean theorem must be \begin{align*}
\end{align*} We have also that the plane $P:3x+2y+6z=1$ is tangent to the sphere $S$. So, must \begin{align*}
\end{align*} The system of the equations $(1)$ and $(2)$, has the solutions $\{c=8,\,R=10\}$ and $\big\{c=\frac{160}{13},\,R=\frac{178}{13}\big\}$. So there are two spheres which satisfy the given conditions.

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