## Precalculus (6th Edition)

$x^2+(y-2)^2=4$ Refer to the graph below.
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.