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

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

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

center

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 1)^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{21}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, - \frac{1}{2}), r = \frac{21}{2}

And the center is:

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

x^{2} + y^{2} - 16 x - 14 y + 31 = 0

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

x^{2} + y^{2} + 2 y - 15 = 0

x^{2} + y^{2} + 4 x - 2 y - 11 = 0

x^{2} - 11 x - 6 y + y^{2} - 11 = 0