What is the center 3x^2+3y^2-17x=0 ?

The center 3x^2+3y^2-17x=0 is (17/6 ,0)

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center 3x^2+3y^2-17x=0

center

3 x^{2} + 3 y^{2} - 17 x = 0

(\frac{17}{6}, 0)

Solution steps

3 x^{2} + 3 y^{2} - 17 x = 0

Rewrite

3 x^{2} + 3 y^{2} - 17 x = 0

in the form of the standard circle equation

(x - \frac{17}{6})^{2} + (y - 0)^{2} = (\frac{17}{6})^{2}

Therefore the circle properties are:

(a, b) = (\frac{17}{6}, 0), r = \frac{17}{6}

And the center is:

(\frac{17}{6}, 0)

(x + 5)^{2} + (y - 2)^{2} = 15

y^{2} + x^{2} = 25

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

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

x^{2} + y^{2} - 4 x - 8 y + 18 = 0