College Algebra, Chapter 2, 2.2, Section 2.2, Problem 68

Show that the equation $x^2 + y^2 + 6y + 2 = 0$ represents a circle. Find the center and radius of the circle.


$
\begin{equation}
\begin{aligned}

x^2 + y^2 + 6y + 2 =& 0
&& \text{Model}
\\
\\
x^2 + y^2 + 6y =& -2
&& \text{Subtract } 2
\\
\\
x^2 + (y^2 + 6y + \underline{ }) =& -2
&& \text{Group terms}
\\
\\
x^2 + (y^2 + 6y + 9) =& -2 + 9
&& \text{Complete the square: add } \left( \frac{6}{2} \right)^2 = 9
\\
\\
x^2 + (y + 3)^2 =& 7
&& \text{Perfect Square}

\end{aligned}
\end{equation}
$


Recall that the general equation for the circle with
circle $(h,k)$ and radius $r$ is..

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

By observation,

The center is at $(0, -3)$ and the radius is $\sqrt{7}$.