What is the center 2x^2+2y^2-4x-8y=10 ?

The center 2x^2+2y^2-4x-8y=10 is (1,2)

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center 2x^2+2y^2-4x-8y=10

center

(1, 2)

Solution steps

2 x^{2} + 2 y^{2} - 4 x - 8 y = 10

Rewrite

2 x^{2} + 2 y^{2} - 4 x - 8 y = 10

in the form of the standard circle equation

(x - 1)^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (1, 2), r = 10

And the center is:

(1, 2)