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    #1
    $\tiny{s6.12.13}$$\textsf{Find an equation of the sphere}\\$ $\textsf{that passes through the point (4,3,-1) and has center (3,8,1)} $ \begin{align}\displaystyle(x-3)^2+(y-8)^2+(z-1)^2&= r^2\\\sqrt{(3-4)^2+(8-3)^2+(1+1)^2}&=r^2\\\sqrt{1+25+4}&=\sqrt{30}^2=30 =r\end{align}$\textit{so far ??}$

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    #2
    What you have is:

    $ \displaystyle r=\sqrt{30}\implies r^2=30$

    I would simply write:

    $ \displaystyle r^2=(3-4)^2+(8-3)^2+(1+1)^2=30$

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    #3 Thread Author
    $\tiny{s6.12.13}$
    $\textsf{Find an equation of the sphere}\\$
    $\textsf{that passes through the point (4,3,-1) and has center (3,8,1)}$
    \begin{align}
    \displaystyle
    (x-3)^2+(y-8)^2+(z-1)^2&= r^2\\
    r^2=(3-4)^2+(8-3)^2+(1+1)^2&=30
    \end{align}

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