Answer
$(x+3)^2+(y+3)^2=61$
Work Step by Step
The midpoint of the segment joining $(3,2)$ and $(-9,-8)$ is the center of the circle $(h,k)$:
$$h=\frac{x_1+x_2}{2}=\frac{3+(-9)}{2}=-3$$ $$k=\frac{y_1+y_2}{2}=\frac{2+(-8)}{2}=-3$$
The distance from the center of the circle to a point on the circle is the radius $r$ of the circle. Finding $r$ using the distance formula:
$$r=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt {(-3-3)^2+(-3-2)^2}=\sqrt {36+25}=\sqrt {61}$$
Using the standard form of the equation of a circle:
$$(x−h)^2+(y−k)^2=r^2$$ $$[x-(-3)]^2+[y-(-3)]^2=(\sqrt {61})^2$$ Simplifying:
$$(x+3)^2+(y+3)^2=61$$
