Answer
$ \ Radius = 3$ and Center at $(2,-2,-1)$
Work Step by Step
Let $(a,b,c)$ be the center of the sphere, $(x,y,z)$ be an arbitrary point of the sphere, and $r$ be the distance between $(a,b,c)$ and $(x,y,z)$. Then, using the Pythagorean Theorem, the standard form of the equation of the sphere can be expressed as:
$r=\sqrt {(x-a)^2+(y-b)^2+(z-c)^2}$
or, $(x-a)^2+(y-b)^2+(z-c)^2=r^2 ...(1)$
We are given
$x^2+y^2+z^2-4x+4y+2z=0$
We complete the square as follows:
$(x^2-4x) +(y^2+4y) + (z^2 +2z) =0 \\ (x^2-4x+4) +(y^2 +4y+4) +(z^2 +2z+1) =0+4+4+1 \\ (x -2)^2+(y +2)^2+(z+1)^2 = 9$
On comparing the above equation with equation (1), we have:
$ \ Radius, r= \sqrt 9=3$ and Center at $(2,-2,-1)$
