Rewrite the equation (2x−2)2 + 4(-y−3)2 = 12 in standard form (x−h)2 + (y−k)2 = r2 and find the center and radius.

For the first term, factor out 2:

$$(2x-2)^2 = (2(x-1))^2 = 2^2(x-1)^2 = 4(x-1)^2$$

For the second term, factor out -1 from $(-y-3)$:

$$(-y-3)^2 = (-(y+3))^2 = (-1)^2(y+3)^2 = (y+3)^2$$

Substitute these back into the equation:

$$4(x-1)^2 + 4(y+3)^2 = 12$$

$$\frac{4(x-1)^2}{4} + \frac{4(y+3)^2}{4} = \frac{12}{4}$$ $$(x-1)^2 + (y+3)^2 = 3$$

Comparing $(x-1)^2 + (y+3)^2 = 3$ with $(x-h)^2 + (y-k)^2 = r^2$:
The center $(h,k)$ is $(1, -3)$.
The radius squared $r^2$ is 3, so the radius $r$ is $\sqrt{3}$.

The radius is $\boxed{\sqrt{3}}$
The coordinates of the center are $\boxed{(1, -3)}$