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

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

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center 4(x^2+y^2-x-2y)=1

center

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - 1)^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 1), r = \frac{3}{2}

And the center is:

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

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

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

9 x^{2} + 9 y^{2} + 18 x - 18 y + 14 = 0

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

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