What is x^2+y^2<= 4x+4y ?

The solution to x^2+y^2<= 4x+4y is Circle with (a,b)=(2,2),r=2sqrt(2)

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x^2+y^2<= 4x+4y

x^{2} + y^{2} \leq 4 x + 4 y

(a, b) = (2, 2), r = 22

Solution steps

x^{2} + y^{2} \leq 4 x + 4 y

Rewrite

x^{2} + y^{2} \leq 4 x + 4 y

in the form of the standard circle equation

(x - 2)^{2} + (y - 2)^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (2, 2), r = 22

What is x^2+y^2<= 4x+4y ?
The solution to x^2+y^2<= 4x+4y is Circle with (a,b)=(2,2),r=2sqrt(2)