Answer
(a) $(h,k)=(3,0)$ and radius $r=2$.
(b) See graph.
(c) $(1,0),(5,0)$.
Work Step by Step
(a) Given $2(x-3)^2+2(y)^2=8\longrightarrow (x-3)^2+(y)^2=4$, we can find the center $(h,k)=(3,0)$ and radius $r=2$.
(b) See graph.
(c) We can find the x-intercept(s) (let y=0) $(1,0),(5,0)$, y-intercept(s) (let x=0) $none$.
