What is the center (x-(-5))^2+(y-(-6))^2= 2/3 ?

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

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center (x-(-5))^2+(y-(-6))^2= 2/3

center

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

(- 5, - 6)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 5, - 6), r = \frac{2}{3}

And the center is:

(- 5, - 6)

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35

3 x^{2} + 3 y^{2} - 18 x - 30 y - 6 = 0

(x - 4)^{2} + (y - 1)^{2} = 218

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

(x + 2)^{2} + (y - 3)^{2} = 5

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