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

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

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center x^2+y^2= 1/4

center

x^{2} + y^{2} = \frac{1}{4}

(0, 0)

Solution steps

x^{2} + y^{2} = \frac{1}{4}

Rewrite

x^{2} + y^{2} = \frac{1}{4}

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = \frac{1}{2}

And the center is:

(0, 0)

x^{2} + y^{2} - 2 x + 6 y - 15 = 0

4 x^{2} - 4 x + 4 y^{2} - 63 = 0

(x - 3)^{2} + (y + 4)^{2} = 25

x^{2} + y^{2} - 8 x + 12 y = - 43

16 x^{2} + 16 y^{2} + 96 x - 40 y = 315