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

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

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

center

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

(- 1, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 1, 2)

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

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

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

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