What is the center x^2+4x+y^2-2y+1=0 ?

The center x^2+4x+y^2-2y+1=0 is (-2,1)

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center x^2+4x+y^2-2y+1=0

center

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

(- 2, 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 1), r = 2

And the center is:

(- 2, 1)

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

(x - 3)^{2} + (y + 5)^{2} = 25

x^{2} + y^{2} - 6 x + 4 y - 3 = 0

(x + 12)^{2} + (y - 9)^{2} = 35

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