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

The solution to 3x^2+3y^2+4y-7=0 is Circle with (a,b)=(0,-2/3),r= 5/3

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

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

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

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

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

(x^{2} - 14 x + 49) + (y^{2} + 8 y + 16) - 25 = 0

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

x^{2} + (y - 3)^{2} = 25

What is 3x^2+3y^2+4y-7=0 ?
The solution to 3x^2+3y^2+4y-7=0 is Circle with (a,b)=(0,-2/3),r= 5/3