Precalculus, Chapter 1, Review Exercises, Section Review Exercises, Problem 22

Determine the center and radius of the circle $(x+2)^2 + y^2 = 9$. Graph the circle. Find the intercepts, if any.

The center of an equation of a circle $(x+2)^2 + y^2 = 9$ is $(-2,0)$ with radius $r = 3$.







To find the $x$-intercepts, we let $y = 0$. Then


$
\begin{equation}
\begin{aligned}

(x+2)^2 + y^2 =& 9
&& \text{Given equation}
\\
(x+2)^2 + 0^2 =& 9
&& y = 0
\\
(x + 2)^2 =& 9
&& \text{Simplify}
\\
x + 2 =& \pm 3
&& \text{Simplify}
\\
x =& -2 \pm 3
&& \text{Solve for } x

\end{aligned}
\end{equation}
$


The $x$-intercepts are $1$ and $-5$.

To find the $y$-intercepts, we let $x = 0$. Then


$
\begin{equation}
\begin{aligned}

(x+2)^2 + y^2 =& 9
&& \text{Given equation}
\\
(0+2)^2 + y^2 =& 9
&& x = 0
\\
4 + y^2 =& 9
&& \text{Apply the Square Root Method}
\\
y =& \pm \sqrt{5}
&& \text{Solve for } y

\end{aligned}
\end{equation}
$


The $y$-intercept is $\pm \sqrt{5}$.