What is 3x^2+3y^2+2x-4y=-1/2 ?

The solution to 3x^2+3y^2+2x-4y=-1/2 is Circle with (a,b)=(-1/3 , 2/3),r=(sqrt(7))/(3sqrt(2))

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3x^2+3y^2+2x-4y=-1/2

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

(a, b) = (- \frac{1}{3}, \frac{2}{3}), r = \frac{7}{3 2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{3}))^{2} + (y - \frac{2}{3})^{2} = (\frac{7}{3 2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, \frac{2}{3}), r = \frac{7}{3 2}

x^{2} - 5 = - y^{2}

(x - 5)^{2} + (y + 2)^{2} = 64

9 (\frac{x - 3}{9})^{2} + 9 (\frac{y + 12}{9})^{2} = 172

(x - 1.5)^{2} + (y + 2.5)^{2} = 1.15

(x + 2)^{2} + (y - 8)^{2} = 4^{2}

What is 3x^2+3y^2+2x-4y=-1/2 ?
The solution to 3x^2+3y^2+2x-4y=-1/2 is Circle with (a,b)=(-1/3 , 2/3),r=(sqrt(7))/(3sqrt(2))