What is the center x^2+y^2+2x+10/3 y-16=0 ?

The center x^2+y^2+2x+10/3 y-16=0 is (-1,-5/3)

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center x^2+y^2+2x+10/3 y-16=0

center

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

(- 1, - \frac{5}{3})

Solution steps

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

Rewrite

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- \frac{5}{3}))^{2} = (\frac{178}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 1, - \frac{5}{3}), r = \frac{178}{3}

And the center is:

(- 1, - \frac{5}{3})

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

x^{2} + y^{2} = 21

x^{2} + y^{2} - 20 x + 36 y + 353 = 0

(x + 1)^{2} + (y - 5)^{2} = 4

(x + 1)^{2} + (y + 3)^{2} = 9

What is the center x^2+y^2+2x+10/3 y-16=0 ?
The center x^2+y^2+2x+10/3 y-16=0 is (-1,-5/3)