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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x - 3)^{2} + (y - 5)^{2} = 1 4^{2}

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

2 x^{2} + 3 x + 2 y^{2} + 7 y = 57.75

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

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

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