What is the center 2x^2+2y^2-17x=0 ?

The center 2x^2+2y^2-17x=0 is (17/4 ,0)

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center 2x^2+2y^2-17x=0

center

2 x^{2} + 2 y^{2} - 17 x = 0

(\frac{17}{4}, 0)

Solution steps

2 x^{2} + 2 y^{2} - 17 x = 0

Rewrite

2 x^{2} + 2 y^{2} - 17 x = 0

in the form of the standard circle equation

(x - \frac{17}{4})^{2} + (y - 0)^{2} = (\frac{17}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{17}{4}, 0), r = \frac{17}{4}

And the center is:

(\frac{17}{4}, 0)

x^{2} + y^{2} - 2 x + 6 y - 6 = 0

(x - 2)^{2} + (y + 3)^{2} = 36

(x - 3)^{2} + (y + 1)^{2} = 25

x^{2} + y^{2} - 4 x - 2 y - 4 = 0

(x - 1)^{2} + (y + 2)^{2} = 16