What is x^2+y^2-4x-4y=0 ?

The solution to x^2+y^2-4x-4y=0 is Circle with (a,b)=(2,2),r=2sqrt(2)

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x^2+y^2-4x-4y=0

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

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

Solution steps

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

Rewrite

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

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

x^{2} = - 4 y

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

x = y^{2} - 4

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

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

What is x^2+y^2-4x-4y=0 ?
The solution to x^2+y^2-4x-4y=0 is Circle with (a,b)=(2,2),r=2sqrt(2)