What is x^2+y^2+4/3 x-4/3 y+4/9 =0 ?

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

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x^2+y^2+4/3 x-4/3 y+4/9 =0

x^{2} + y^{2} + \frac{4}{3} x - \frac{4}{3} y + \frac{4}{9} = 0

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

Solution steps

x^{2} + y^{2} + \frac{4}{3} x - \frac{4}{3} y + \frac{4}{9} = 0

Rewrite

x^{2} + y^{2} + \frac{4}{3} x - \frac{4}{3} y + \frac{4}{9} = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

3 x^{2} - 9 x + 3 y^{2} + 3 y = \frac{39}{2}

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

x^{2} + y^{2} + 4 x - 8 y + 11 = 0

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

2 x^{2} + 2 y^{2} + 16 x + 12 y + 48 = 0

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