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

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

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

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

(a, b) = (\frac{1}{2}, 0), r = \frac{3}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 0), r = \frac{3}{2}

3 x^{2} - 5 x + 3 y^{2} + 5 y - 8 = 0

(x + 4)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} + 6 x + 8 x = - 16

x^{2} + y^{2} - 10 x - 6 y + 25 = 0

x^{2} + y^{2} > 1

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