Answer
(a) The circle has a center (-2,2) and a radius of 3 units.
(b) See below.
(c) There are two x-intercepts: $(-4.24,0) \text{ and } (0.24,0)$.
There are two y-intercepts: $(0,-0.24) \text{ and } (0,4.24)$.
Work Step by Step
First, we need to make sure that the equation is in standard form $(x-h)^2+(y-k)^2=r^2$
In this case, we need to complete the square to achieve the standard form:
$x^2+y^2+4x-4y-1=0$
$x^2+4x+y^2-4y=1$
$x^2+4x+(\frac{4}{2})^2+y^2-4y+(\frac{4}{2})^2=1+(\frac{4}{2})^2+(\frac{4}{2})^2$
$(x+2)^2+(y-2)^2=1+4+4$
$(x+2)^2+(y-2)^2=9$
$(x+2)^2+(y-2)^2=3^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x+2)^2+(0-2)^2=9$
$(x+2)^2+4=9$
$(x+2)^2=5$
$\sqrt{(x+2)^2}=\sqrt5$
There are two x-intercepts:
$x_1+2=-\sqrt5\rightarrow x_1=-\sqrt5-2\approx-4.24$
$x_2+2=\sqrt5\rightarrow x_2=\sqrt5-2\approx0.24$
The y-intercepts are all points of a graph when x=0:
$(0+2)^2+(y-2)^2=9$
$4+(y-2)^2=9$
$(y-2)^2=5$
$\sqrt{(y-2)^2}=\sqrt{5}$
There are two y-intercepts:
$y_1-2=-\sqrt5\rightarrow y_1=-\sqrt5+2\approx-0.24$
$y_2-2=\sqrt5\rightarrow y_2=\sqrt5+2\approx4.24$