Answer
(a). $(h,k)=(3,-1), r=1$.
(b). See graph.
(c). $(3,0)$.
Work Step by Step
(a). $x^2+y^2-6x+2y+9=0 \longrightarrow (x-3)^2+(y+1)^2=1$, thus $(h,k)=(3,-1), r=1$.
(b). See graph.
(c). For x-intercept(s), let $y=0$, we get $(3,0)$. For y-intercept(s), let $x=0$, we get $none$.
