Answer
Center $(-2,0,2)$ and radius $2 \sqrt 2$
Work Step by Step
As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ )
where $(a,b,c)$ represents center and radius of the sphere is $r$
Now, $(x+2)^2+y^2+(z-2)^2=8$
$\implies (x-(-2))^2+(y-0)^2+(z-2)^2=(\sqrt 8)^2$
$ \implies (x-(-2))^2+(y-0)^2+(z-2)^2=( 2 \sqrt 2)^2$
Compare this equation with the standard equation of sphere.
Thus, Center $(-2,0,2)$ and radius $2 \sqrt 2$
