Answer
a) The center is $(h,k)=(3,0)$ and the radius is $r=2$
b) The graph is attached below.
c) There are two $x$-intercepts $(5,0)$ and $(1,0)$
There are no $y$-intercepts.
Work Step by Step
$(a)$
If we start with the standard form of the equation, we can easily get the center and the radius of the circle by finding the constants $h, k$ and $r$:
$(x-h)^2+(y-k)^2=r^2$
Here, $2(x-3)^2+2y^2=8$.
Divide both sides by $2$ to obtain:
$(x-3)^2+y^2=4$
$(x-3)^2+(y-0)^2=2^2$
Therefore the center is $(h,k)=(3,0)$ and the radius is $r=2$.
$(b)$
To graph the circle, plot the center $3, 0)$.
Then, with a radius of $2$ units, plot the following points:
point $2$ units directly above the center: $(3, 2)$
point $2$ units directly below the center: $(3, -2)$
point $2$ units to the left of the center: $(1, 0)$
point $2$ units to the right of the center: $(5, 0)$
Finally, connect these four points using a smooth curve. Refer to the graph above.
$(c)$
The $x$- and $y$-intercepts can be calculated by plugging $0$ into $y$ and $x$, respectively.
$x$-intercepts:
$(x-3)^2+y^2=4$
$(x-3)^2+0^2=4$
$(x-3)^2=4$
$x-3=\pm2$
$x=3\pm2$
$x_1=5$
$x_2=1$
Therefore there are two $x$-intercepts $(5,0)$ and $(1,0)$
$y$-intercepts:
$(x-3)^2+y^2=4$
$(0-3)^2+y^2=4$
$9+y^2=4$
$y^2=-5$ this is not possible, therefore there is no $y$-intercepts.
