Answer
(a) The circle has a center (-2,0) and a radius of about 0.71 units.
(b) See below.
(c) There are two x-intercepts: $(-2.71,0) \text{ and } (-1.29,0)$. There are no y-intercepts.
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:
$2x^2+2y^2+8x+7=0$
$2(x^2+y^2+4x+3.5)=0$
$x^2+y^2+4x+3.5=0$
$x^2+4x+y^2=-3.5$
$x^2+4x+(\frac{4}{2})^2+y^2=-3.5+(\frac{4}{2})^2$
$(x+2)^2+y^2=-3.5+4$
$(x+2)^2+y^2=0.5$
$(x+2)^2+y^2=(\sqrt{0.5})^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x+2)^2+0^2=0.5$
$(x+2)^2=0.5$
$\sqrt{(x+2)^2}=\sqrt{0.5}$
There are two x-intercepts:
$x_1+2=-\sqrt{0.5}\rightarrow x_1=-\sqrt{0.5}-2\approx-2.71$
$x_2+2=\sqrt{0.5}\rightarrow x_2=\sqrt{0.5}-2\approx-1.29$
The y-intercepts are all points of a graph when x=0:
$(0+2)^2+y^2=0.5$
$4+y^2=0.5$
$y^2=-3.5$
$\sqrt{y^2}=\sqrt{-3.5}$
Taking the square root of a negative number isn't possible; therefore, the graph doesn't intersect with the y-axis at any point.
