What is x^2+y^2<= x ?

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

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x^2+y^2<= x

x^{2} + y^{2} \leq x

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

Solution steps

x^{2} + y^{2} \leq x

Rewrite

x^{2} + y^{2} \leq x

in the form of the standard circle equation

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

Therefore the circle properties are:

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

4 x^{2} + 4 y^{2} + 4 x - 27 = 0

x^{2} + y^{2} + 6 x - 8 y + 12 = 0

4 (x^{2} - 2 x + 1) + 4 (y^{2} - y + 0) = 11 + 1 + 0

x^{2} + y^{2} + x + y = 85 - 7

x^{2} + y^{2} - 10 x + 12 y - 83 - 0 = 0

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