Answer
$x^{2}+(y-2)^{2}=4$
Work Step by Step
The circle with center $C(h,k)$ and radus $a$ has the general equation
$( x-h)^{2}+(y-k)^{2}=a^{2}.$
Here, $\qquad $
$( x-0)^{2}+(y-2)^{2}=4.$
$x^{2}+(y-2)^{2}=4$
x-intercepts (y=0):
$x^{2}+(0-2)^{2}=4$
$x^{2}+4=4$
$x^{2}=0$
$ x=0\qquad$... $\quad(0,0)$
y-intercepts (x=0):
$0+(y-2)^{2}=4$
$y-2=\pm 2$
$y=2\pm 2$
$y=0,\quad y=4\qquad (0,0),(0,4)$