What is 3x^2+3y^2+6x+9y-12=0 ?

The solution to 3x^2+3y^2+6x+9y-12=0 is Circle with (a,b)=(-1,-3/2),r=(sqrt(29))/2

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3x^2+3y^2+6x+9y-12=0

3 x^{2} + 3 y^{2} + 6 x + 9 y - 12 = 0

(a, b) = (- 1, - \frac{3}{2}), r = \frac{29}{2}

Solution steps

3 x^{2} + 3 y^{2} + 6 x + 9 y - 12 = 0

Rewrite

3 x^{2} + 3 y^{2} + 6 x + 9 y - 12 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{29}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 1, - \frac{3}{2}), r = \frac{29}{2}

x^{2} + y^{2} - 4 y - 6 y + 9 = 0

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

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

x^{2} + y^{2} - 2 \geq 0

u^{2} + y^{2} = 146

What is 3x^2+3y^2+6x+9y-12=0 ?
The solution to 3x^2+3y^2+6x+9y-12=0 is Circle with (a,b)=(-1,-3/2),r=(sqrt(29))/2