What is the center x^2+10x+y^2-16-11=0 ?

The center x^2+10x+y^2-16-11=0 is (-5,0)

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center x^2+10x+y^2-16-11=0

center

x^{2} + 10 x + y^{2} - 16 - 11 = 0

(- 5, 0)

Solution steps

x^{2} + 10 x + y^{2} - 16 - 11 = 0

Rewrite

x^{2} + 10 x + y^{2} - 16 - 11 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 0)^{2} = (213)^{2}

Therefore the circle properties are:

(a, b) = (- 5, 0), r = 213

And the center is:

(- 5, 0)

x^{2} + y^{2} + 10 x - 8 y + 25 = 0

4 x^{2} + 4 y^{2} = 144

(x + 1)^{2} + y^{2} = 246

(x - 4)^{2} + (y + 1^{2})^{2} = 9

(x + \frac{9}{4})^{2} = \frac{13}{2} - (y - 8)^{2}

What is the center x^2+10x+y^2-16-11=0 ?
The center x^2+10x+y^2-16-11=0 is (-5,0)