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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 9)^{2} + (y + 5)^{2} = 49

x^{2} + y^{2} + 16 x - 16 y + 124 = 0

(x - 1)^{2} + (y + 5)^{2} = 10

(x + 5)^{2} + (y - 4)^{2} = 9

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

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