Answer
(a) The circle has a center (-1,1) and a radius of $\sqrt2$ units.
(b) See below.
(c) There are two x-intercepts: (-2,0) and (0,0). There are two y-intercepts: (0,0) 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 tidy the equation up a bit:
$3(x+1)^2+3(y-1)^2=6$
$(3(x+1)^2+3(y-1)^2)/3=6/3$
$(x+1)^2+(y-1)^2=2$
$(x+1)^2+(y-1)^2=\sqrt2^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x+1)^2+(0-1)^2=2$
$(x+1)^2+1=2$
$(x+1)^2=1$
$\sqrt{(x+1)^2}=\sqrt1$
There are two x-intercepts:
$x_1+1=-1$
$x_1=-2$
$x_2+1=1$
$x_2=0$
The y-intercepts are all points of a graph when x=0:
$(0+1)^2+(y-1)^2=2$
$1+(y-1)^2=2$
$(y-1)^2=1$
$\sqrt{(y-1)^2}=\sqrt{1}$
There are two y-intercepts:
$y_1-1=-1$
$y_1=0$
$y_2-1=1$
$y_2=2$
