Answer
(a) The circle has a center (0.5,-1) and a radius of 0.5 units long.
(b) See below.
(c) There are no x-intercepts. There is only one y-intercept: (0,-1).
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:
$x^2+y^2-x+2y+1=0$
$x^2-x+y^2+2y=-1$
$x^2-x+(\frac{1}{2})^2+y^2+2y+(\frac{2}{2})^2=-1+(\frac{1}{2})^2+(\frac{2}{2})^2$
$(x-0.5)^2+(y+1)^2=-1+0.25+1$
$(x-0.5)^2+(y+1)^2=0.25$
$(x-0.5)^2+(y+1)^2=0.5^2$
Now we can find the center (h,k) and radius.
The x-intercepts are all points of a graph when y=0:
$(x-0.5)^2+(0+1)^2=0.25$
$(x-0.5)^2+1=0.25$
$(x-0.5)^2=-0.75$
$\sqrt{(x-0.5)^2}=\sqrt{-0.75}$
Taking the square root of a negative number is not possible; therefore, the graph doesn't intercept the x-axis at any point.
The y-intercepts are all points of a graph when x=0:
$(0-0.5)^2+(y+1)^2=0.25$
$0.25+(y+1)^2=0.25$
$(y+1)^2=0$
$\sqrt{(y+1)^2}=\sqrt{0}$
There is one y-intercept:
$y+1=0\rightarrow y=-1$