Answer
$x^2+(y-2)^2=4$
See graph.
Work Step by Step
A circle has the form:
$(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$
Our center is $(0,2)$. Since the circle is tangent to the $x$-axis, the radius must be $2$ (vertical distance from x-axis to the center). So we have:
$(x-0)^2+(y-2)^2=2^2=4$
$x^2+(y-2)^2=4$