Answer
$x^2+(y-2)^2=4$
Refer to the graph below.
Work Step by Step
RECALL:
A circle whose equation is of the form $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and has a radius of $r$.
The given circle has its center at $(0, 2)$ and tangent to the x-axis.
Since the center is $2$ units away from the x-axis, then the circle has a radius of $2$ units.
Therefore, the equation of the circle is:
$(x-0)^2+(y-2)^2=2^2
\\x^2+(y-2)^2=4$
Refer to the graph in the answer part above.
