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

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

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

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

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

Solution steps

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

Rewrite

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

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}

x^{2} + y^{2} + 2 x - 6 y + 8 = 0

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x^{2} + y^{2} - 10 x + 16 y + 25 = 0

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12 x^{2} + 12 y^{2} + 12 x - 24 y + 3 = 0

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