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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x + 5)^{2} + (y + 3)^{2} = 18

2 x^{2} + 8 x + 2 y^{2} - 6 y = 18

x^{2} - 110 + y^{2} = 1

x^{2} + y^{2} + 3 x - y - 1 = 0

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

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