What is the center 9x^2+9y^2-6x+24y=19 ?

The center 9x^2+9y^2-6x+24y=19 is (1/3 ,-4/3)

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center 9x^2+9y^2-6x+24y=19

center

(\frac{1}{3}, - \frac{4}{3})

Solution steps

9 x^{2} + 9 y^{2} - 6 x + 24 y = 19

Rewrite

9 x^{2} + 9 y^{2} - 6 x + 24 y = 19

in the form of the standard circle equation

(x - \frac{1}{3})^{2} + (y - (- \frac{4}{3}))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{3}, - \frac{4}{3}), r = 2

And the center is:

(\frac{1}{3}, - \frac{4}{3})

What is the center 9x^2+9y^2-6x+24y=19 ?
The center 9x^2+9y^2-6x+24y=19 is (1/3 ,-4/3)