What is the center y^2+x^2+24x+10y+160=0 ?

The center y^2+x^2+24x+10y+160=0 is (-12,-5)

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center y^2+x^2+24x+10y+160=0

center

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

(- 12, - 5)

Solution steps

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

Rewrite

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

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} = 9

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

y^{2} + 4 x - 20 - 2 y = - x^{2}

(x - 1)^{2} + (y - 3)^{2} = 16

(x + 3)^{2} + (y - 3)^{2} = 8

What is the center y^2+x^2+24x+10y+160=0 ?
The center y^2+x^2+24x+10y+160=0 is (-12,-5)