## Calculus 8th Edition

center,$(h,k)=(0,-3)$ and $r= \sqrt 7$
An equation of the circle with center $(h,k)$ and radius $r$ is given as: $(x-h)^{2}+(y-k)^{2}=r^{2}$ ...(1) Given: $x^{2}+ y^{2}+6y +2 = 0$ The above equation can be written in the standard equation of the circle as follows: $(x-0)^{2}+ (y-(-3))^{2} = (\sqrt 7)^{2}$ Compare it with equation (1), we have Hence, center,$(h,k)=(0,-3)$ and $r= \sqrt 7$