What is x^2+y^2-4x+2y-8=0 ?

The solution to x^2+y^2-4x+2y-8=0 is Circle with (a,b)=(2,-1),r=sqrt(13)

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x^2+y^2-4x+2y-8=0

x^{2} + y^{2} - 4 x + 2 y - 8 = 0

(a, b) = (2, - 1), r = 13

Solution steps

x^{2} + y^{2} - 4 x + 2 y - 8 = 0

Rewrite

x^{2} + y^{2} - 4 x + 2 y - 8 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 1))^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (2, - 1), r = 13

0.5 9^{2} = (x - (- 0.02))^{2} + (y - 0.78)^{2}

(x + \frac{23}{5})^{2} + (y - \frac{2}{5})^{2} = \frac{373}{5}

4 x^{2} + 4 y^{2} - 8 x + 16 y + 11 = 0

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

(x + 7)^{2} + (y + 4)^{2} = 49

What is x^2+y^2-4x+2y-8=0 ?
The solution to x^2+y^2-4x+2y-8=0 is Circle with (a,b)=(2,-1),r=sqrt(13)