What is the center 3x^2+3y^2-12x-2=0 ?

The center 3x^2+3y^2-12x-2=0 is (2,0)

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center 3x^2+3y^2-12x-2=0

center

3 x^{2} + 3 y^{2} - 12 x - 2 = 0

(2, 0)

Solution steps

3 x^{2} + 3 y^{2} - 12 x - 2 = 0

Rewrite

3 x^{2} + 3 y^{2} - 12 x - 2 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = (\frac{14}{3})^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = \frac{14}{3}

And the center is:

(2, 0)

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35

3 x^{2} + 3 y^{2} - 18 x - 30 y - 6 = 0

(x - 4)^{2} + (y - 1)^{2} = 218

4 x^{2} + 4 y^{2} - x + 2 y - 11 = 0