Answer
The circle has a center (1,-2) and a radius of $\sqrt5$ units long.
There are two x-intercepts: (0,0) and (2,0).
There are two y-intercepts: (0,-4) and (0,0).
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:
$3x^2+3y^2-6x+12y=0$
$3(x^2+y^2-2x+4y)=0$
$x^2+y^2-2x+4y=0$
$x^2-2x+y^2+4y=0$
$x^2-2x+(\frac{2}{2})^2+y^2+4y+(\frac{4}{2})^2=(\frac{2}{2})^2+(\frac{4}{2})^2$
$(x-1)^2+(y+2)^2=1+4$
$(x-1)^2+(y+2)^2=5$
$(x-1)^2+(y+2)^2=(\sqrt5)^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x-1)^2+(0+2)^2=5$
$(x-1)^2+4=5$
$(x-1)^2=1$
$\sqrt{(x-1)^2}=\sqrt1$
There are two x-intercepts:
$x_1-1=-1\rightarrow x_1=0$
$x_2-1=1\rightarrow x_2=2$
The y-intercepts are all points of a graph when x=0:
$(0-1)^2+(y+2)^2=5$
$1+(y+2)^2=5$
$(y+2)^2=4$
$\sqrt{(y+2)^2}=\sqrt{4}$
There are two y-intercepts:
$y_1+2=-2\rightarrow y_1=-4$
$y_2+2=2\rightarrow y_2=0$
