What is the center 3x^2+3y^2+2x-4y-1=0 ?

The center 3x^2+3y^2+2x-4y-1=0 is (-1/3 , 2/3)

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center 3x^2+3y^2+2x-4y-1=0

center

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

(- \frac{1}{3}, \frac{2}{3})

Solution steps

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

Rewrite

3 x^{2} + 3 y^{2} + 2 x - 4 y - 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, \frac{2}{3}), r = \frac{2 2}{3}

And the center is:

(- \frac{1}{3}, \frac{2}{3})

x^{2} + y^{2} = 36

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

4 x^{2} + 4 y^{2} = 25

x^{2} + y^{2} = 64

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

What is the center 3x^2+3y^2+2x-4y-1=0 ?
The center 3x^2+3y^2+2x-4y-1=0 is (-1/3 , 2/3)