Answer
(a) The circle has a center (3,-2) and a radius of 5 units.
(b) See below.
(c) There are two x-intercepts: $(-1.58,0) \text{ and } (7.58,0)$.
There are two y-intercepts: $(0,-6) \text{ and } (0,2)$.
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:
$2x^2+2y^2-12x+8y-24=0$
$2(x^2+y^2-6x+4y-12)=0$
$x^2+y^2-6x+4y-12=0$
$x^2-6x+y^2+4y=12$
$x^2-6x+(\frac{6}{2})^2+y^2+4y+(\frac{4}{2})^2=12+(\frac{6}{2})^2+(\frac{4}{2})^2$
$(x-3)^2+(y+2)^2=12+9+4$
$(x-3)^2+(y+2)^2=25$
$(x-3)^2+(y+2)^2=5^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x-3)^2+(0+2)^2=25$
$(x-3)^2+4=25$
$(x-3)^2=21$
$\sqrt{(x-3)^2}=\sqrt{21}$
There are two x-intercepts:
$x_1-3=-\sqrt{21}\rightarrow x_1=-\sqrt{21}+3\approx-1.58$
$x_2-3=\sqrt{21}\rightarrow x_2=\sqrt{21}+3\approx7.58$
The y-intercepts are all points of a graph when x=0:
$(0-3)^2+(y+2)^2=25$
$9+(y+2)^2=25$
$(y+2)^2=16$
$\sqrt{(y+2)^2}=\sqrt{16}$
There are two y-intercepts:
$y_1+2=-4\rightarrow y_1=-6$
$y_2+2=4\rightarrow y_2=2$