Answer
(a). $(h,k)=(3,-2), r=5$.
(b). See graph.
(c). $(3\pm\sqrt {21},0)$, $(0,-6),(0,2)$.
Work Step by Step
(a). $x^2+y^2-6x+4y=12\longrightarrow (x-3)^2+(y+2)^2=12+9+4=25$, thus $(h,k)=(3,-2), r=5$.
(b). See graph.
(c). For x-intercept(s), let $y=0$, we get $(3\pm\sqrt {21},0)$, for y-intercept(s), let $x=0$, we get $(0,-2\pm4)$ or $(0,-6),(0,2)$.
