Answer
$(x+7)^{2}$ + $(y-6)^{2}$ = 4
Work Step by Step
In this problem, $h$ is -7, $k$ is 6, and $r$ is 2. Given the equation $(x-h)^{2}$ + $(y-k)^{2}$ = $r^{2}$, the equation of this circle would be $(x-(-7))^{2}$ + $(y-(6))^{2}$ = $2^{2}$, or $(x+7)^{2}$ + $(y-6)^{2}$ = $4$.
