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

The solution to 3x^2+3y^2-3x-3y-9=0 is Circle with (a,b)=(1/2 , 1/2),r=sqrt(7/2)

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

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

(a, b) = (\frac{1}{2}, \frac{1}{2}), r = \frac{7}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = (\frac{7}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, \frac{1}{2}), r = \frac{7}{2}

(x - 6)^{2} + (y + 2)^{2} = 16

x^{2} - 8 x + 16 + y^{2} - 4 y + 4 = (34)^{2}

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

3 x^{2} + 3 y^{2} - 6 x - y + 1 = 0

x^{2} - 4 x + y^{2} + 12 y = 16

What is 3x^2+3y^2-3x-3y-9=0 ?
The solution to 3x^2+3y^2-3x-3y-9=0 is Circle with (a,b)=(1/2 , 1/2),r=sqrt(7/2)