Answer
$(x+1)^{2}+(y-5)^{2}=10$
.
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 $
$( x-(-1))^{2}+(y-5)^{2}=10$
$(x+1)^{2}+(y-5)^{2}=10$
x-intercepts (y=0):
$(x+1)^{2}+(0-5)^{2}=10$
$(x+1)^{2}+25=-10$
$(x+1)^{2}=-15\qquad$... $\quad$ none
y-intercepts (x=0):
$(0+1)^{2}+(y-5)^{2}=10$
$(y-5)^{2}=10-1$
$y-5=\pm 3$
$y=5\pm 3$
$y=2,\quad y=8\qquad (0,8),(0,2)$
