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

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

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

center

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

(\frac{1}{2}, - 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - 1), r = \frac{1}{2}

And the center is:

(\frac{1}{2}, - 1)

(x - 2)^{2} + (y - 3)^{2} = 49

(x - 5)^{2} + (y + 7)^{2} = 4

3 x^{2} + 3 y^{2} - 12 x - 2 = 0

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35