What is 3x^2+3y^2+2x-6y-4=0 ?

The solution to 3x^2+3y^2+2x-6y-4=0 is Circle with (a,b)=(-1/3 ,1),r=(sqrt(22))/3

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3x^2+3y^2+2x-6y-4=0

3 x^{2} + 3 y^{2} + 2 x - 6 y - 4 = 0

(a, b) = (- \frac{1}{3}, 1), r = \frac{22}{3}

Solution steps

3 x^{2} + 3 y^{2} + 2 x - 6 y - 4 = 0

Rewrite

3 x^{2} + 3 y^{2} + 2 x - 6 y - 4 = 0

in the form of the standard circle equation

(x - (- \frac{1}{3}))^{2} + (y - 1)^{2} = (\frac{22}{3})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, 1), r = \frac{22}{3}

(x - 7.5)^{2} + (y - 7.84)^{2} = 0.5 6^{2}

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

x^{2} - 8 x + y^{2} - 6 y + 25 = 1

x^{2} + y^{2} + 6 x - 2 x - 2 y + 1 = 0

x^{2} + y^{2} - 6 x - 2 y = 15

What is 3x^2+3y^2+2x-6y-4=0 ?
The solution to 3x^2+3y^2+2x-6y-4=0 is Circle with (a,b)=(-1/3 ,1),r=(sqrt(22))/3