#### Answer

$y=2+\sqrt{4-x^2}$
With domain: $-2\leq x \leq2$

#### Work Step by Step

We need to solve for $y$:
$x^2+(y-2)^2=4$
$(y-2)^2=4-x^2$
$y-2=\pm\sqrt{4-x^2}$
$y=2\pm\sqrt{4-x^2}$
Since we need the top half of the circle, we use:
$y=2+\sqrt{4-x^2}$
With domain: $-2\leq x \leq2$