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If $A$ and $B$ be the points $(3,4,5)$ and $(-1,3,-7)$, respectively, find the equation of the set of points $P$ such that $P A^2+P B^2=k^2$, where $k$ is a constant.
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Verified Answer
Let the coordinate of $P$ be $(x, y, z)$
The given points are $A(3,4,5)$ and $B(-1,3,-7)$.
$\begin{aligned}
&P A^2+P B^2 \\
&=\left[(x-3)^2+(y-4)^2+(z-5)^2\right]+\left[(x+1)^2\right. \\
&\left.\quad+(y-3)^2+(z+7)^2\right] \\
&=\left[x^2+y^2+z^2-6 x-8 y-10 z+9+16+25\right] \\
&\quad+\left[x^2+y^2+z^2+2 x-6 y+14 z+1+9+49\right]
\end{aligned}$
or $2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+50+59=k^2$
or $2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+109-k^2=0$
The given points are $A(3,4,5)$ and $B(-1,3,-7)$.
$\begin{aligned}
&P A^2+P B^2 \\
&=\left[(x-3)^2+(y-4)^2+(z-5)^2\right]+\left[(x+1)^2\right. \\
&\left.\quad+(y-3)^2+(z+7)^2\right] \\
&=\left[x^2+y^2+z^2-6 x-8 y-10 z+9+16+25\right] \\
&\quad+\left[x^2+y^2+z^2+2 x-6 y+14 z+1+9+49\right]
\end{aligned}$
or $2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+50+59=k^2$
or $2\left(x^2+y^2+z^2\right)-4 x-14 y+4 z+109-k^2=0$
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