why is

x^2+y^2+z^2=z

a sphere centered in (x,y,z)=(0,0,1/2)?

It doesn't. The center is (0,0,1). And it's not at (0,0,0) because the equation equals Z. What you have is

(x^2) + (y^2) + (z^2 - z) = 0

So you need to find the points where each term equals zero in order to find the center of the sphere. The x and y terms are obviously zero when x and y are zero. So when you have (0,0,z) as your center and you want to find out how to satisfy that equation, you have

0^2 + 0^2 + (z^2 - z) = 0
z^2 - z = 0
z^2 = z

The only value of z that satisfies this is z = 1.

Edit: just realized my reasoning is HORRIBLE. But damn it all, I don't feel like figuring it out. I'll just strike it out so you can witness my fail and that I'm owning up to it.

This post was edited by bentherdonethat on Dec 9 2011 02:53am

Because your equation is equivalent to

(x - 0)^2 + (y - 0)^2 + (z - 1/2)^2=1/2^2

which is the equation for a sphere with radius 1/2 with center at (0, 0, 1/2)

Yep, gotta complete the square of ' z^2 - z ' then you should have the form (x-c1)^2+(y-c2)^2+(z-c3)^2=r^2