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

The solution to x^2-2x+y^2-4y-4=0 is Circle with (a,b)=(1,2),r=3

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

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

(a, b) = (1, 2), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 2), r = 3

(x - 1)^{2} = 27 - (y + 1)^{2}

(x - 5)^{2} + (y - 6)^{2} = 20

x^{2} - 6 x + y^{2} + 10 y + 9 = 0

x^{2} + y^{2} + 12 x - 8 y + 16 = 0

(x - 4)^{2} + (y - 20)^{2} = 592^{2}

What is x^2-2x+y^2-4y-4=0 ?
The solution to x^2-2x+y^2-4y-4=0 is Circle with (a,b)=(1,2),r=3