What is x^2+4x+y^2+8y+16=0 ?

The solution to x^2+4x+y^2+8y+16=0 is Circle with (a,b)=(-2,-4),r=2

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x^2+4x+y^2+8y+16=0

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

(a, b) = (- 2, - 4), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = 2

(x^{2} + (- 10) x + 25) + (y^{2} + 9 y + \frac{81}{4}) = - 1 + 25 + \frac{81}{4}

x^{2} + 6 x + y^{2} - 2 y - 26 = 0

x^{2} + y^{2} + 9 x - 10 = 0

2 y - y^{2} - 6 x - x^{2} = 9

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

What is x^2+4x+y^2+8y+16=0 ?
The solution to x^2+4x+y^2+8y+16=0 is Circle with (a,b)=(-2,-4),r=2