What is x^2+4x+y^2+8y-5=0 ?

The solution to x^2+4x+y^2+8y-5=0 is Circle with (a,b)=(-2,-4),r=5

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x^2+4x+y^2+8y-5=0

x^{2} + 4 x + y^{2} + 8 y - 5 = 0

(a, b) = (- 2, - 4), r = 5

Solution steps

x^{2} + 4 x + y^{2} + 8 y - 5 = 0

Rewrite

x^{2} + 4 x + y^{2} + 8 y - 5 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = 5

x^{2} - 2 x + y^{2} - 15 = 0

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

2 x^{2} + 3 x + 2 y^{2} + 7 y = 52.75

(x - 5)^{2} + (y - 2)^{2} = 49

(x - 15)^{2} + [y - (- 20)]^{2} = (9)^{2}

What is x^2+4x+y^2+8y-5=0 ?
The solution to x^2+4x+y^2+8y-5=0 is Circle with (a,b)=(-2,-4),r=5