What is the center 2(x-3)^2+2y^2=8 ?

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

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center 2(x-3)^2+2y^2=8

center

2 (x - 3)^{2} + 2 y^{2} = 8

(3, 0)

Solution steps

2 (x - 3)^{2} + 2 y^{2} = 8

Rewrite

2 (x - 3)^{2} + 2 y^{2} = 8

in the form of the standard circle equation

(x - 3)^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (3, 0), r = 2

And the center is:

(3, 0)

x^{2} + 16 x + y^{2} - 12 y = 0

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

16 x^{2} + 16 y^{2} = 320

x^{2} + y^{2} - 14 x + 12 y = 36

(x + 3)^{2} + (y - 5)^{2} = 5