What is the center 3x^2+3y^2+4y-7=0 ?

The center 3x^2+3y^2+4y-7=0 is (0,-2/3)

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center 3x^2+3y^2+4y-7=0

center

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

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

Solution steps

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

Rewrite

3 x^{2} + 3 y^{2} + 4 y - 7 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{2}{3}))^{2} = (\frac{5}{3})^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{2}{3}), r = \frac{5}{3}

And the center is:

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

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x^{2} + y^{2} - 8 x - 10 y + 40 = 0