## Algebra and Trigonometry 10th Edition

Center: $(0,-12)$ Radius: $r=2\sqrt 6$
The equation of a circle in standard form: $(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius $x^2+(y+12)^2=24$ $(x-0)^2+[y-(-12)]^2=(\sqrt {24})^2$ Center: $(0,-12)$ Radius: $r=\sqrt {24}=2\sqrt 6$