College Algebra, Chapter 2, 2.2, Section 2.2, Problem 72

Show that the equation $\displaystyle x^2 + y^2 + \frac{1}{2} x + 2y + \frac{1}{16} = 0$ represents a circle. Find the center and radius of the circle.


$
\begin{equation}
\begin{aligned}

x^2 + y^2 + \frac{1}{2} x + 2y + \frac{1}{16} =& 0
&& \text{Model}
\\
\\
\left( x^2 + \frac{1}{2} x + \underline{ } \right) + (y^2 + 2y + \underline{ }) =& \frac{-1}{16}
&& \text{Group terms and subtract } \frac{1}{16}
\\
\\
\left( x^2 + \frac{1}{2}x + \underline{\frac{1}{16}} \right) + (y^2 + 2y + \underline{1}) =& \frac{-1}{16} + \frac{1}{16} + 1
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{1}{2}}{2} \right)^2 = \frac{1}{16} \text{ and } \left( \frac{2}{2} \right)^2 = 1
\\
\\
\left( x + \frac{1}{4} \right)^2 + (y + 1)^2 =& 1
&& \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 $\displaystyle \left( \frac{-1}{4}, -1 \right)$ and the radius is $1$.

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