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

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

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

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

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

Solution steps

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

Rewrite

x^{2} + 2 x + y^{2} - 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 - 6)^{2} + (y - (- 2))^{2} = 25

x^{2} + y^{2} - 8 x - 1 y - 74 = 0

0.5 y^{2} = - 0.5 x^{2} + 12.5

p^{2} + q^{2} = 1

x^{2} + y^{2} + 14 x - 6 y - 6 = 0

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