What is a^2-2a+b^2-2b>= 0 ?

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

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a^2-2a+b^2-2b>= 0

a^{2} - 2 a + b^{2} - 2 b \geq 0

(a, b) = (1, 1), r = 2

Solution steps

a^{2} - 2 a + b^{2} - 2 b \geq 0

Rewrite

a^{2} - 2 a + b^{2} - 2 b \geq 0

in the form of the standard circle equation

(a - 1)^{2} + (b - 1)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (1, 1), r = 2

(x + 3)^{2} + (y - 1)^{2} = 121

x^{2} + (y + 5)^{2} = 16

(x - 2)^{2} + (y - 1)^{2} = 8

4 x^{2} + 32 x + 4 y^{2} + 20 y = 0

(x - 4)^{2} + (y - 3)^{2} = 5^{2}

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