Answer
(a) The circle has a center (-2,-1) and a radius of 5 units.
(b) See below.
(c) There are two x-intercepts: $(-6.9,0) \text{ and } (2.9,0)$.
There are two y-intercepts: $(0,-5.58) \text{ and } (0,3.58)$.
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:
$x^2+y^2+4x+2y-20=0$
$x^2+4x+y^2+2y=20$
$x^2+4x+(\frac{4}{2})^2+y^2+2y+(\frac{2}{2})^2=20+(\frac{4}{2})^2+(\frac{2}{2})^2$
$(x+2)^2+(y+1)^2=20+4+1$
$(x+2)^2+(y+1)^2=25$
$(x+2)^2+(y+1)^2=5^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x+2)^2+(0+1)^2=25$
$(x+2)^2+1=25$
$(x+2)^2=24$
$\sqrt{(x+2)^2}=\sqrt{24}$
There are two x-intercepts:
$x_1+2=-\sqrt{24}\rightarrow x_1=-\sqrt{24}-2\approx-6.9$
$x_2+2=\sqrt{24}\rightarrow x_2=\sqrt{24}-2\approx2.9$
The y-intercepts are all points of a graph when x=0:
$(0+2)^2+(y+1)^2=25$
$4+(y+1)^2=25$
$(y+1)^2=21$
$\sqrt{(y+1)^2}=\sqrt{21}$
There are two y-intercepts:
$y_1+1=-\sqrt{21}\rightarrow y_1=-\sqrt{21}-1\approx-5.58$
$y_2+1=\sqrt{21}\rightarrow y_2=\sqrt{21}-1\approx3.58$