What is x^2+y^2-4x-2y-5=0 ?

The solution to x^2+y^2-4x-2y-5=0 is Circle with (a,b)=(2,1),r=sqrt(10)

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x^2+y^2-4x-2y-5=0

x^{2} + y^{2} - 4 x - 2 y - 5 = 0

(a, b) = (2, 1), r = 10

Solution steps

x^{2} + y^{2} - 4 x - 2 y - 5 = 0

Rewrite

x^{2} + y^{2} - 4 x - 2 y - 5 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 1)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (2, 1), r = 10

2 x^{2} + 2 y^{2} + 2 x - 6 y - 7 = 0

(x - 4)^{2} + (y - 6)^{2} = 8

(x - 4)^{2} + (y - 5)^{2} = 16

16 x^{2} + 16 y^{2} + 96 x + 40 y = 315

(x - 4)^{2} + (y - 5)^{2} = 25

What is x^2+y^2-4x-2y-5=0 ?
The solution to x^2+y^2-4x-2y-5=0 is Circle with (a,b)=(2,1),r=sqrt(10)