What is 3x^2+3y^2+12x+12y=4 ?

The solution to 3x^2+3y^2+12x+12y=4 is Circle with (a,b)=(-2,-2),r=(2sqrt(7))/(sqrt(3))

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3x^2+3y^2+12x+12y=4

3 x^{2} + 3 y^{2} + 12 x + 12 y = 4

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

Solution steps

3 x^{2} + 3 y^{2} + 12 x + 12 y = 4

Rewrite

3 x^{2} + 3 y^{2} + 12 x + 12 y = 4

in the form of the standard circle equation

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

Therefore the circle properties are:

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

2 x^{2} + 2 y^{2} - 12 x + 4 y - 12 = 0

(x - 3)^{2} + (y - 5)^{2} = 144

x^{2} + y^{2} - 16 x + 40 y + 67 = 0

x^{2} + y^{2} + 8 x + 2 y + 9 = 0

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

What is 3x^2+3y^2+12x+12y=4 ?
The solution to 3x^2+3y^2+12x+12y=4 is Circle with (a,b)=(-2,-2),r=(2sqrt(7))/(sqrt(3))