Answer
$(x-2)^{2}+(y+1)^{2}=16$
Work Step by Step
We know that the standard form of the equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^{2}+(y-k)^{2}=r^{2}$.
Therefore, we can plug in the given values for $h$, $k$, and $r$ and simplify.
$(x-2)^{2}+(y-(-1))^{2}=(4)^{2}$
$(x-2)^{2}+(y+1)^{2}=16$
