Answer
(a) The circle has a center (-2,0) and a radius of 2 units.
(b) See below.
(c) There are two x-intercepts: (-4,0) and (0,0).
There is only one y-intercept: (0,0).
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:
$2x^2+8x+2y^2=0$
$2(x^2+4x+y^2)=0$
$x^2+4x+y^2=0$
$x^2+4x+(\frac{4}{2})^2+y^2=(\frac{4}{2})^2$
$(x+2)^2+y^2=4$
$(x+2)^2+y^2=2^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x+2)^2+0^2=4$
$(x+2)^2=4$
$\sqrt{(x+2)^2}=\sqrt4$
There are two x-intercepts:
$x_1+2=-2\rightarrow x_1=-4$
$x_2+2=2\rightarrow x_2=0$
The y-intercepts are all points of a graph when x=0:
$(0+2)^2+y^2=4$
$4+y^2=4$
$y^2=0$
$\sqrt{y^2}=\sqrt{0}$
There is only one y-intercept:
$y=0$
