Question

Show that the equation represents a circle, and find the center and radius of the circle.

$x^2+y^2+6y+2=0$

Short answer

The Centre of the circle is$(0,-3)$

The radius of the circle is$\sqrt{7}$

Step-by-step solution

Step 1. Determining general Equation of circle.

An equation of the circle with center$(h,k)$and radius is

${(x-h)}^2+{(y-k)}^2=r^2$

This is called the standard form for the equation of the circle.

Step 2. Representing the equation is a circle. 

$\begin{array}{c}{x}^2+y^2+6y+2=0\\ \left(x^2\right)+(y^2+6y)=-2(\mathrm{Group}\mathrm{terms})\\ x^2+(y^2+6y+9)=-2+9(\mathrm{Complete}\mathrm{the}\mathrm{square}\mathrm{by}\mathrm{adding}9\mathrm{to}\mathrm{both}\mathrm{side})\\ x^2+{\left(y+3\right)}^2=7(\mathrm{Factor}\mathrm{and}\mathrm{simplify})\end{array}$

Step 3. Determining the radius and center of the circle.

${(x-0)}^2+{(y+3)}^2=7$

Comparing the equation with the standard equation of a circle

$h=0,k=-3,r=\sqrt{7}$

So, the Centre of the circle$(0,-3)$ is and the radius of the circle is$\sqrt{7}$