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 Post subject: Equation of sphere
PostPosted: Wed Dec 16, 2015 11:52 am 
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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 sphere
PostPosted: Wed Sep 14, 2016 8:26 am 
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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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Then, by Pythagorean theorem must be \begin{align*}
{\rm{MK}_1}^2={\rm{KK}_1}^2+{\rm{KM}}^2\quad&\Rightarrow\quad R^2=c^2+r^2\\
&\Rightarrow\quad R=\sqrt{c^2+36}\qquad(1)
\end{align*} We have also that the plane $P:3x+2y+6z=1$ is tangent to the sphere $S$. So, must \begin{align*}
{\rm{d}}({\rm{K}_1},P)=R\quad&\Rightarrow\quad\dfrac{|3\cdot5+2\cdot4+6c-1|}{\sqrt{3^2+2^2+6^2}}=R\\
&\Rightarrow\quad\dfrac{|22+6c|}{7}=R\qquad(2).
\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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