What is the center x^2+y^2+10x+6y+18=0 ?

The center x^2+y^2+10x+6y+18=0 is (-5,-3)

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center x^2+y^2+10x+6y+18=0

center

x^{2} + y^{2} + 10 x + 6 y + 18 = 0

(- 5, - 3)

Solution steps

x^{2} + y^{2} + 10 x + 6 y + 18 = 0

Rewrite

x^{2} + y^{2} + 10 x + 6 y + 18 = 0

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 3))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 3), r = 4

And the center is:

(- 5, - 3)

(x - 4)^{2} + (y - 15)^{2} = 169

(x - 2)^{2} + (y + 5)^{2} = 16

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

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

(x + 5)^{2} + y^{2} = 9

What is the center x^2+y^2+10x+6y+18=0 ?
The center x^2+y^2+10x+6y+18=0 is (-5,-3)