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

The center x^2+10x+y^2-16y-11=0 is (-5,8)

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

center

(- 5, 8)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 8)^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (- 5, 8), r = 10

And the center is:

(- 5, 8)

What is the center x^2+10x+y^2-16y-11=0 ?
The center x^2+10x+y^2-16y-11=0 is (-5,8)