What is the center 2x^2+2y^2-8x+7=0 ?

The center 2x^2+2y^2-8x+7=0 is (2,0)

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center 2x^2+2y^2-8x+7=0

center

(2, 0)

Solution steps

2 x^{2} + 2 y^{2} - 8 x + 7 = 0

Rewrite

2 x^{2} + 2 y^{2} - 8 x + 7 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = \frac{1}{2}

And the center is:

(2, 0)