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

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

(- 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)

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

x^{2} + y^{2} + 8 y = 33

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

x^{2} + y^{2} + 6 x - 2 y + 1 = 0

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)