Answer
$ \ Radius, r= \sqrt 1=1$ and Center at $(-1,0,1)$
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:
$x^2+y^2+z^2+2x-2z=-1$
We complete the square as: follows:
$(x^2+2x) +(y^2) + (z^2-2z) =-1 \\ (x^2+2x+1) +y^2 +(z^2-2z+1) =-1+1+1 \\ (x+1)^2+(y-0)^2+(z-1)^2 =1$
On comparing the above equation with equation (1), we have:
$ \ Radius, r= \sqrt 1=1$ and Center at $(-1,0,1)$