What is 2x-x^2-y^2>= 0 ?

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

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2x-x^2-y^2>= 0

2 x - x^{2} - y^{2} \geq 0

(a, b) = (1, 0), r = 1

Solution steps

2 x - x^{2} - y^{2} \geq 0

Rewrite

2 x - x^{2} - y^{2} \geq 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 0), r = 1

16 x^{2} + 16 y^{2} + 8 x - 32 y = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{25} = 1

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

x^{2} + y^{2} - 2 x - 4 y - 15 = 0

4 x^{2} + 4 y^{2} - 24 y + 51 = 16 x

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