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

(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)

4 x^{2} + 4 y^{2} - 16 x + 40 y + 76 = 0

x^{2} + y^{2} \leq 9

x^{2} + 10 x + y^{2} - 16 - 11 = 0

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

4 x^{2} + 4 y^{2} = 144