What is 9x^2+9y^2-6x+12y+4=0 ?

The solution to 9x^2+9y^2-6x+12y+4=0 is Circle with (a,b)=(1/3 ,-2/3),r= 1/3

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9x^2+9y^2-6x+12y+4=0

9 x^{2} + 9 y^{2} - 6 x + 12 y + 4 = 0

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

Solution steps

9 x^{2} + 9 y^{2} - 6 x + 12 y + 4 = 0

Rewrite

9 x^{2} + 9 y^{2} - 6 x + 12 y + 4 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 10 x - 16 y - 55 = 0

(x + 8)^{2} + (y + 5)^{2} = 256

x^{2} - 16 x + y^{2} - 36 = 0

(x - \frac{7}{4})^{2} + (y - \frac{3}{4})^{2} = \frac{169}{16}

x^{2} + y^{2} + 6 x - 22 y + 120 = 0

What is 9x^2+9y^2-6x+12y+4=0 ?
The solution to 9x^2+9y^2-6x+12y+4=0 is Circle with (a,b)=(1/3 ,-2/3),r= 1/3