What is a^2+b^2<= 2a+1 ?

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

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a^2+b^2<= 2a+1

a^{2} + b^{2} \leq 2 a + 1

(a, b) = (1, 0), r = 2

Solution steps

a^{2} + b^{2} \leq 2 a + 1

Rewrite

a^{2} + b^{2} \leq 2 a + 1

in the form of the standard circle equation

(a - 1)^{2} + (b - 0)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (1, 0), r = 2

(x - 7)^{2} + (y - 7)^{2} = \frac{4}{9}

x^{2} + y^{2} - 10 x + 6 y + 18 = 0

(x + 3)^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} - 6 x + 2 y + 1 = 0

x^{2} + y^{2} - 4 x + 10 y + 23 = 0

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