S6.12.13 Find an equation of the sphere

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SUMMARY

The equation of a sphere that passes through the point (4,3,-1) and has its center at (3,8,1) is derived using the standard formula for a sphere's equation, which is \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\). Here, the center coordinates are \(h=3\), \(k=8\), and \(l=1\). The radius \(r\) is calculated as \(r = \sqrt{30}\), leading to \(r^2 = 30\). Thus, the final equation of the sphere is \((x-3)^2 + (y-8)^2 + (z-1)^2 = 30\).

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karush
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$\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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What you have is:

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

I would simply write:

$$r^2=(3-4)^2+(8-3)^2+(1+1)^2=30$$
 
$\tiny{s6.12.13}$
$\textsf{Find an equation of the sphere}\\$
$\textsf{that passes through the point (4,3,-1) and has center (3,8,1)}$
\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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