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

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

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

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

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

Solution steps

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

Rewrite

x^{2} + y^{2} - 4 x - 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^{2} + (y + 2)^{2} = 36

(x + 5)^{2} + (y + 4)^{2} = 9

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

(x - 2)^{2} + (y - 3)^{2} = 25

\frac{2 x ^{2} + 2 y ^{2} - 3 x + 5 y - 3}{2} = 0

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