Answer
$(x+2)^2+(y-3)^2=16$
Work Step by Step
RECALL:
The standard form of a circle's equation is given as:
$(x-h)^2+(y-k)^2=r^2$
where $r$ = radius and $(h, k)$ is the center.
Thus, a circle with center at $(-2, 3)$ and $r=4$ has the equation:
$[(x-(-2)]^2+(y-3)^2=4^2
\\(x+2)^2+(y-3)^2=16$