Answer
center,$(h,k)=(0,-3)$ and $r= \sqrt 7$
Work Step by Step
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 $r= \sqrt 7$
