What is the center 9x^2+12x+9y^2-18y-23=0 ?

The center 9x^2+12x+9y^2-18y-23=0 is (-2/3 ,1)

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center 9x^2+12x+9y^2-18y-23=0

center

9 x^{2} + 12 x + 9 y^{2} - 18 y - 23 = 0

(- \frac{2}{3}, 1)

Solution steps

9 x^{2} + 12 x + 9 y^{2} - 18 y - 23 = 0

Rewrite

9 x^{2} + 12 x + 9 y^{2} - 18 y - 23 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- \frac{2}{3}, 1)

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

(x - 1)^{2} + (y + 3)^{2} = 25

(x + 14)^{2} + (y - 11)^{2} = 16

x^{2} + y^{2} - 6 x + 10 y + 9 = 0

x^{2} + y^{2} + 6 x + 8 y + 24 = 0

What is the center 9x^2+12x+9y^2-18y-23=0 ?
The center 9x^2+12x+9y^2-18y-23=0 is (-2/3 ,1)