Show that $${ x^{2}+y^{2}+z^{2}+2x+4y-2z=10 }$$

is the equation of a sphere. Find its center and radius.

Solution We begin by writing the equation as $$\begin{equation*} (x^{2}+2x)+(y^{2}+4y)+(z^{2}-2z)=10 \end{equation*}$$

and then we complete the square three times. The result is $$\begin{eqnarray*} (x^{2}+2x+1)+(y^{2}+4y+4)+(z^{2}-2z+1) &=&10+1+4+1 \\[4pt] (x+1)^{2}+(y+2)^{2}+(z-1)^{2} &=&16 \end{eqnarray*}$$

This is the equation of a sphere with radius $4$ and center at $(-1,-2, 1)$.