Answer
Center: $(-2,0,2 )$ and radius: $\sqrt 8$
Work Step by Step
We know that the standard equation of a sphere is written as:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\ \ \ $ (1)
Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere.
Since, we have $x^2+y^2+z^2+4x-4z=0$
or, $x^2+4x+y^2+z^2-4z=0$
or, $(x - 2)^2+y^2+(z - 2)^2=4+4$
or, $(x - 2)^2+y^2+(z - 2)^2=8$
or, or, $(x -(- 2))^2+y^2+(z - 2)^2=(\sqrt 8)^2$
Compare the above equation with equation (1), we get
Center: $(-2,0,2 )$ and radius: $\sqrt 8$
