Precalculus, Chapter 1, Review Exercises, Section Review Exercises, Problem 24

Determine the center and radius of the circle $x^2 + y^2 + 4x - 4y - 1 = 0$. Graph the circle. Find the intercepts, if any.

We complete the square in both $x$ and $y$ to put the equation in standard form


$
\begin{equation}
\begin{aligned}

x^2 + y^2 + 4x - 4y - 1 =& 0
&& \text{Given equation}
\\
(x^2 + 4x) + (y^2 - 4y) =& 1
&& \text{Group the equation in terms of $x$ and $y$. And put the consistent on the right side of the equation}
\\
(x^2 + 4x + 4) + (y^2 - 4y + 4) =& 1 + 4 + 4
&& \text{Complete the square: add } \left( \frac{4}{2} \right)^2 = 4 \text{ and } \left( \frac{4}{2} \right)^2 = 4 \text{ on both sides of the equation}
\\
(x + 2)^2 + (y - 2)^2 =& 9
&& \text{Factor}

\end{aligned}
\end{equation}
$


We recognize this equation as the standard form of the equation of a circle with $r=3$ and center $(-2,2)$







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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


The $x$-intercepts are $-2 + \sqrt{5}$ and $-2- \sqrt{5}$.

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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


The $y$-intercepts are $2 + \sqrt{5}$ and $2 - \sqrt{5}$.