What is x^2+y^2-4x+2y-1=0 ?

The solution to x^2+y^2-4x+2y-1=0 is Circle with (a,b)=(2,-1),r=sqrt(6)

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x^2+y^2-4x+2y-1=0

x^{2} + y^{2} - 4 x + 2 y - 1 = 0

(a, b) = (2, - 1), r = 6

Solution steps

x^{2} + y^{2} - 4 x + 2 y - 1 = 0

Rewrite

x^{2} + y^{2} - 4 x + 2 y - 1 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 1))^{2} = (6)^{2}

Therefore the circle properties are:

(a, b) = (2, - 1), r = 6

x^{2} + y^{2} + 14 x - 12 y = - 81

(x - 3)^{2} + (y - 4)^{2} = 1^{2}

4 x^{(2)} + 4 y^{(2)} - 24 x + 16 y + 51 = 0

x^{2} + y^{2} + 12 x - 10 y = - 45

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

What is x^2+y^2-4x+2y-1=0 ?
The solution to x^2+y^2-4x+2y-1=0 is Circle with (a,b)=(2,-1),r=sqrt(6)