Answer
(a) The circle has a center (3,0) and a radius of 2 units.
(b) See below.
(c) There are two x-intercepts: (1,0) and (5,0). 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 tidy the equation up a bit:
$2(x-3)^2+2y^2=8$
$(2(x-3)^2+2y^2)/2=8/2$
$(x-3)^2+y^2=4$
$(x-3)^2+y^2=2^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x-3)^2+0^2=4$
$(x-3)^2=4$
$\sqrt{(x-3)^2}=\sqrt4$
There are two x-intercepts:
$x_1-3=-2$
$x_1=1$
$x_2-3=2$
$x_2=5$
The y-intercepts are all points of a graph when x=0:
$(0-3)^2+y^2=4$
$(-3)^2+y^2=4$
$9+y^2=4$
$y^2=-5$
$\sqrt{y^2}=\sqrt{-5}$
Taking the square root of a negative number isn't possible; therefore, this circle doesn't intersect the y-axis at any point.
