Answer
(a). $(h,k)=(\frac{1}{2},-1), r=\frac{1}{2}$.
(b). See graph.
(c). $(0,-1)$.
Work Step by Step
(a). $x^2+y^2-x+2y+1=0 \longrightarrow (x-\frac{1}{2})^2+(y+1)^2=\frac{1}{4}$, thus $(h,k)=(\frac{1}{2},-1), r=\frac{1}{2}$.
(b). See graph.
(c). For x-intercept(s), let $y=0$, we get $none$, for y-intercept(s), let $x=0$, we get $(0,-1)$.
