What is 3x^2+3y^2+6y-20=0 ?

The solution to 3x^2+3y^2+6y-20=0 is Circle with (a,b)=(0,-1),r=sqrt(23/3)

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3x^2+3y^2+6y-20=0

3 x^{2} + 3 y^{2} + 6 y - 20 = 0

(a, b) = (0, - 1), r = \frac{23}{3}

Solution steps

3 x^{2} + 3 y^{2} + 6 y - 20 = 0

Rewrite

3 x^{2} + 3 y^{2} + 6 y - 20 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = (\frac{23}{3})^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = \frac{23}{3}

x^{2} + y^{2} > 9

x^{2} + y^{2} - 4 x + 2 y + 1 = 0

(x - 6)^{2} + (y + 36)^{2} = 25

x^{2} + y^{2} - 7 y + 8 = 0

x^{2} + 6 x + y^{2} - 10 y = 135

What is 3x^2+3y^2+6y-20=0 ?
The solution to 3x^2+3y^2+6y-20=0 is Circle with (a,b)=(0,-1),r=sqrt(23/3)