Answer
$(x-2)^2+(y+3)^2=13$
Work Step by Step
The middlepoint between $(0,0)$ and $(4,-6)$ is the center of the circle:
$\frac{(0,0)+(4,-6)}{2}=(2,-3)$, that is: $h=2$, $k=-3$
The distance from the center to an endpoint is the radius:
$r=\sqrt {(2-0)^2+(-3-0)^2}=\sqrt {13}$
Equation of a circle:
$(x−h)^2+(y−k)^2=r^2$ (standard form)
$(x-2)^2+[y-(-3)]^2=(\sqrt {13})^2$
$(x-2)^2+(y+3)^2=13$
