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)

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

(x - 4)^{2} + (y + 1)^{2} = 18

(x - 4)^{2} + (y + 1)^{2} = 16

x^{2} + y^{2} - 6 x - 12 y - 55 = 0

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

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)