What is x^2+y^2-2x+8y+1=0 ?

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

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x^2+y^2-2x+8y+1=0

x^{2} + y^{2} - 2 x + 8 y + 1 = 0

(a, b) = (1, - 4), r = 4

Solution steps

x^{2} + y^{2} - 2 x + 8 y + 1 = 0

Rewrite

x^{2} + y^{2} - 2 x + 8 y + 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 4), r = 4

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

3 x^{2} + 3 y^{2} + 30 x - 12 y - 60 = 0

2 x^{2} + 2 y^{2} - 16 x + 8 y - 10 = 0

x^{2} + x + (p (x)^{2} - 7 p (x)) = 0

4 x^{2} + 4 y^{2} - 32 x + 8 y - 32 = 0

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