What is the center (x+8/3)^2+y^2=1 ?

The center (x+8/3)^2+y^2=1 is (-8/3 ,0)

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center (x+8/3)^2+y^2=1

center

(x + \frac{8}{3})^{2} + y^{2} = 1

(- \frac{8}{3}, 0)

Solution steps

(x + \frac{8}{3})^{2} + y^{2} = 1

Rewrite

(x + \frac{8}{3})^{2} + y^{2} = 1

in the form of the standard circle equation

(x - (- \frac{8}{3}))^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- \frac{8}{3}, 0), r = 1

And the center is:

(- \frac{8}{3}, 0)

(x - 3)^{2} + (y + 4)^{2} = \frac{16}{9}

x^{2} + y^{2} + 4 y - 5 = 0

(x + 2)^{2} + (y - 3)^{2} = 15

(x + 5)^{2} + (y + 2)^{2} = 29

x^{2} + y^{2} - 16 x + 12 y + 64 = 0