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

The center x^2+y^2+2x+10y+18=0 is (-1,-5)

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

center

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

(- 1, - 5)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 5))^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 5), r = 22

And the center is:

(- 1, - 5)

x^{2} + (y - 1)^{2} = 9

(x - 11)^{2} + (y + 2)^{2} = 16

(x - \frac{3}{2})^{2} + (y - \frac{3}{2})^{2} = \frac{9}{4}

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

2 x^{2} + 2 y^{2} + 6 x - 8 y + 12 = 0

What is the center x^2+y^2+2x+10y+18=0 ?
The center x^2+y^2+2x+10y+18=0 is (-1,-5)