What is 3x^2+3y^2-12y-8=0 ?

The solution to 3x^2+3y^2-12y-8=0 is Circle with (a,b)=(0,2),r=(2sqrt(5))/(sqrt(3))

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3x^2+3y^2-12y-8=0

3 x^{2} + 3 y^{2} - 12 y - 8 = 0

(a, b) = (0, 2), r = \frac{2 5}{3}

Solution steps

3 x^{2} + 3 y^{2} - 12 y - 8 = 0

Rewrite

3 x^{2} + 3 y^{2} - 12 y - 8 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 2)^{2} = (\frac{2 5}{3})^{2}

Therefore the circle properties are:

(a, b) = (0, 2), r = \frac{2 5}{3}

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

- x^{2} - 6 x = z^{2} + 2 z + 1

x^{2} + y^{2} + 2 x + 12 y - 27 = 0

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

x^{2} + y^{2} + 20 x - 6 y = - 108

What is 3x^2+3y^2-12y-8=0 ?
The solution to 3x^2+3y^2-12y-8=0 is Circle with (a,b)=(0,2),r=(2sqrt(5))/(sqrt(3))