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 $\pm \sqrt{5}$.