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

The solution to 3x^2+4x+3y^2-2y=0 is Circle with (a,b)=(-2/3 , 1/3),r=(sqrt(5))/3

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

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

(a, b) = (- \frac{2}{3}, \frac{1}{3}), r = \frac{5}{3}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{2}{3}))^{2} + (y - \frac{1}{3})^{2} = (\frac{5}{3})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{2}{3}, \frac{1}{3}), r = \frac{5}{3}

(x - 2)^{2} + (y - 1)^{2} = 5^{2}

\frac{( x + 2 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

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

(x - 5)^{2} + (y - 4)^{2} = 3^{2}

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

What is 3x^2+4x+3y^2-2y=0 ?
The solution to 3x^2+4x+3y^2-2y=0 is Circle with (a,b)=(-2/3 , 1/3),r=(sqrt(5))/3