What is x^2+y^2> 1/2 ?

The solution to x^2+y^2> 1/2 is Circle with (a,b)=(0,0),r=sqrt(1/2)

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x^2+y^2> 1/2

x^{2} + y^{2} > \frac{1}{2}

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

Solution steps

x^{2} + y^{2} > \frac{1}{2}

Rewrite

x^{2} + y^{2} > \frac{1}{2}

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}

x^{2} - 8 x + y^{2} - 4 y + 10 = 0

x^{2} + x + y^{2} = 16

x^{2} + y^{2} - \frac{608}{79} x - \frac{1386}{79} y - \frac{5640}{79} = 0

x^{(2)} + y^{(2)} + 4 x - 6 y + 12 = 0

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

What is x^2+y^2> 1/2 ?
The solution to x^2+y^2> 1/2 is Circle with (a,b)=(0,0),r=sqrt(1/2)