Answer
$(x+4)^2+(y-3)^2=106$
Work Step by Step
A circle has the form:
$(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$
We are told that our circle has center $(-4,3)$ (thus $h=-4,k=3$) and passes through the point $(5,8)$. So:
$(x+4)^2+(y-3)^2=r^2$
We plug in $x=5,y=8$:
$(5+4)^2+(8-3)^2=r^2$
$9^2+5^2=r^2$
$81+25=r^2$
$106=r^2$
$r=\sqrt{106}$
Thus the equation is:
$(x+4)^2+(y-3)^2=106$