What is the center 3x^2+3y^2+6x-5y=0 ?

The center 3x^2+3y^2+6x-5y=0 is (-1, 5/6)

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center 3x^2+3y^2+6x-5y=0

center

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

(- 1, \frac{5}{6})

Solution steps

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

Rewrite

3 x^{2} + 3 y^{2} + 6 x - 5 y = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{5}{6})^{2} = (\frac{61}{6})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{5}{6}), r = \frac{61}{6}

And the center is:

(- 1, \frac{5}{6})

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

2 x^{2} + 2 y^{2} + 10 x = 2 y + 7

(x - 2)^{2} + (y + 4)^{2} = 25

x^{2} + y^{2} - 4 x - 8 y - 5 = 0

x^{2} + (y - 5)^{2} = 36

What is the center 3x^2+3y^2+6x-5y=0 ?
The center 3x^2+3y^2+6x-5y=0 is (-1, 5/6)