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

3 x^{2} + 3 y^{2} = y + 2

x^{2} + (y - 115)^{2} = 11025

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

16 (x^{2} + y^{2}) = 16 (14 x + 6 y + 6)

(x + 4)^{2} + (y - 2)^{2} = \frac{( 8 ^{2} )}{( 5 )}

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)