Question 397311
How could I find the center (h,k) and radius r of this equation: 2(x-3)^2 + 2y^2=8 ?

standard equation of a circle with center (h,k) and radius r:
(x - h)^2 + (y - k)^2 = r^2


2(x - 3)^2 + 2y^2 = 8
(x - 3)^2 + y^2 = 4
(x - 3)^2 + (y - 0)^2 = 4
center is (3,0), radius = 2