Answer
$(x+\sqrt{3})^{2}+(y+2)^{2}=4$
.
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 $
$( x-(-\sqrt{3}))^{2}+(y-(-2))^{2}=2^{2}$
$(x+\sqrt{3})^{2}+(y+2)^{2}=4$
x-intercepts (y=0):
$(x+\sqrt{3})^{2}+(0+2)^{2}=4$
$(x+\sqrt{3})^{2}+4=4$
$(x+\sqrt{3})^{2}=0$
$ x=-\sqrt{3}\qquad$... $\quad (-\sqrt{3},0) $
y-intercepts (x=0):
$(0+\sqrt{3})^{2}+(y+2)^{2}=4$
$3+(y+2)^{2}=4$
$(y+2)^{2}=1$
$y+2=\pm 1$
$y=-2\pm 1$
$y=-1,\quad y=-3\qquad (0,-1),(0,-3)$
