Show that \[ { x^{2}+y^{2}+z^{2}+2x+4y-2z=10 } \]

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

Completing the square is discussed in Appendix A.1, pp. A-2 to A-3.

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 \((-1,-2, 1)\).