What is the center x^2+y^2-2/3 x-5/3 y-5/9 =0 ?

The center x^2+y^2-2/3 x-5/3 y-5/9 =0 is (1/3 , 5/6)

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center x^2+y^2-2/3 x-5/3 y-5/9 =0

center

(\frac{1}{3}, \frac{5}{6})

Solution steps

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

Rewrite

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

in the form of the standard circle equation

(x - \frac{1}{3})^{2} + (y - \frac{5}{6})^{2} = (\frac{7}{6})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{3}, \frac{5}{6}), r = \frac{7}{6}

And the center is:

(\frac{1}{3}, \frac{5}{6})

What is the center x^2+y^2-2/3 x-5/3 y-5/9 =0 ?
The center x^2+y^2-2/3 x-5/3 y-5/9 =0 is (1/3 , 5/6)