Answer
(a) The circle has a center (-0.5,-0.5) and a radius of 1 unit.
(b) See below.
(c) There are two x-intercepts: (-1.37,0) and (0.37,0).
There are two y-intercepts: (0,-1.37) and (0,0.37).
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:
$x^2+y^2+x+y-0.5=0$
$x^2+x+y^2+y=0.5$
$x^2+x+(\frac{1}{2})^2+y^2+y+(\frac{1}{2})^2=0.5+(\frac{1}{2})^2+(\frac{1}{2})^2$
$(x+0.5)^2+(y+0.5)^2=0.5+0.25+0.25$
$(x+0.5)^2+(y+0.5)^2=1$
$(x+0.5)^2+(y+0.5)^2=1^2$
Now we can find the center (h,k) and the radius.
The x-intercepts are all points of a graph when y=0:
$(x+0.5)^2+(0+0.5)^2=1$
$(x+0.5)^2+0.25=1$
$(x+0.5)^2=0.75$
$\sqrt{(x+0.5)^2}=\sqrt{0.75}$
There are two x-intercepts:
$x_1+0.5=-\sqrt{0.75}\rightarrow x_1=-\sqrt{0.75}-0.5\approx-1.37$
$x_2+0.5=\sqrt{0.75}\rightarrow x_2=\sqrt{0.75}-0.5\approx0.37$
The y-intercepts are all points of a graph when x=0:
$(0+0.5)^2+(y+0.5)^2=1$
$0.25+(y+0.5)^2=1$
$(y+0.5)^2=0.75$
$\sqrt{(y+0.5)^2}=\sqrt{0.75}$
There are two y-intercepts:
$y_1+0.5=-\sqrt{0.75}\rightarrow y_1=-\sqrt{0.75}-0.5\approx-1.37$
$y_2+0.5=\sqrt{0.75}\rightarrow y_2=\sqrt{0.75}-0.5\approx0.37$
