What is the center x^2+4x+10y+y^2+3=0 ?

The center x^2+4x+10y+y^2+3=0 is (-2,-5)

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center x^2+4x+10y+y^2+3=0

center

x^{2} + 4 x + 10 y + y^{2} + 3 = 0

(- 2, - 5)

Solution steps

x^{2} + 4 x + 10 y + y^{2} + 3 = 0

Rewrite

x^{2} + 4 x + 10 y + y^{2} + 3 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 5))^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 5), r = 26

And the center is:

(- 2, - 5)

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

x^{2} + y^{2} - 6 x + 5 = 0

(x - 1)^{2} + (y + 4)^{2} = 36

x^{2} + y^{2} + 14 x - 12 y + 4 = 0

(x + 2)^{2} + (y - 4)^{2} = 25

What is the center x^2+4x+10y+y^2+3=0 ?
The center x^2+4x+10y+y^2+3=0 is (-2,-5)