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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

5 x^{2} + 5 y^{2} + 20 x + 30 y - 310 = 0

x^{2} + y^{2} = \frac{- x - y + 2}{2}

(X + 7)^{2} + (y - 1)^{2} = 9

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

(x - 3)^{2} + (y + 1)^{2} = 49

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