What is the center x^2+24x+y^2+10y=-160 ?

The center x^2+24x+y^2+10y=-160 is (-12,-5)

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center x^2+24x+y^2+10y=-160

center

x^{2} + 24 x + y^{2} + 10 y = - 160

(- 12, - 5)

Solution steps

x^{2} + 24 x + y^{2} + 10 y = - 160

Rewrite

x^{2} + 24 x + y^{2} + 10 y = - 160

in the form of the standard circle equation

(x - (- 12))^{2} + (y - (- 5))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 12, - 5), r = 3

And the center is:

(- 12, - 5)

(x + 2)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} + 14 x - 16 y + 64 = 0

2 x^{2} - 2 x + 2 y^{2} + 4 y = - 1

4 x^{2} + 4 y^{2} - 4 x + 2 y + 1 = 0

x^{2} + 2 x + y^{2} - 3 = 6

What is the center x^2+24x+y^2+10y=-160 ?
The center x^2+24x+y^2+10y=-160 is (-12,-5)