Answer
A circle with center at $C(2,-2)$ and radius $2\sqrt{2}$
.
Work Step by Step
Gather terms containing x and y in separate parentheses
$(x^{2}-4x)+(y^{2}+4y)=0$
Complete squares:
$x^{2}-4x+4-4=(x-2)^{2}-4$
$y^{2}+4y+4-4=(y+2)^{2}-4$
The equation becomes
$(x-2)^{2}-4+(y+2)^{2}-4=0$
$(x-2)^{2}+(y+2)^{2}=8$
A circle with center at $C(2,-2)$ and radius $2\sqrt{2}$
x-intercepts $:\quad$ $\left[\begin{array}{l}
(x-2)^{2}+(0+2)^{2}=8\\
(x-2)^{2}+4=8\\
(x-2)^{2}=4\\
x-2=\pm 2 \\
x=2\pm 2\\
(0,0),\quad(4,0)
\end{array}\right] $
y-intercepts $:\quad \left[\begin{array}{l}
(0-2)^{2}+(y+2)^{2}=8\\
4+(y+2)^{2}=8\\
(y+2)^{2}=4\\
y+2=\pm 2\\
y=-2\pm 2\\
y=0,\quad y=-4\\
(0,-4),\quad(0,0)
\end{array}\right] $
