Answer
(a) The circle has a center (0,2) and a radius of 2 units.
(b) See below.
(c) The only x-intercept is (0,0). There are two y-intercepts: (0,0) and (0,4).
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:
$3x^2+3y^2-12y=0$
$3(x^2+y^2-4y)=0$
$x^2+y^2-4y=0$
$x^2+y^2-4y+(\frac{4}{2})^2=(\frac{4}{2})^2$
$x^2+(y-2)^2=4$
$x^2+(y-2)^2=2^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$x^2+(0-2)^2=4$
$x^2+4=4$
$x^2=0$
$\sqrt{x^2}=\sqrt0$
There is one x-intercept:
$x=0$
The y-intercepts are all points of a graph when x=0:
$0^2+(y-2)^2=4$
$(y-2)^2=4$
$\sqrt{(y-2)^2}=\sqrt{4}$
There are two y-intercepts:
$y_1-2=-2\rightarrow y_1=0$
$y_2-2=2\rightarrow y_2=4$
