Answer
$ \ radius=r= \sqrt 4=2$ and center is at $(-1,1,0)$
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 the formula:
$x^2+y^2+z^2+2x-2y=2$
We complete the square and rewrite the equation as follows:
$(x^2+2x) +(y^2-2y) +z^2=2 \\ (x^2+2x+1) +(y^2-2y+1) +z^2 =2+1+1 \\ (x+1)^2+(y-1)^2+(z-0)^2 =4$
On comparing the above equation with equation (1), we have:
$ \ Radius=r= \sqrt 4=2$ and Center is at $(-1,1,0)$
