What is x^2+y^2+4x+8y-36=0 ?

The solution to x^2+y^2+4x+8y-36=0 is Circle with (a,b)=(-2,-4),r=2sqrt(14)

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x^2+y^2+4x+8y-36=0

x^{2} + y^{2} + 4 x + 8 y - 36 = 0

(a, b) = (- 2, - 4), r = 214

Solution steps

x^{2} + y^{2} + 4 x + 8 y - 36 = 0

Rewrite

x^{2} + y^{2} + 4 x + 8 y - 36 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 4))^{2} = (214)^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = 214

x^{2} + y^{2} - 6 x - 8 y - 22 = 0

x^{2} + y^{2} - 9 x - 2 y + 17 = 0

x^{2} + y^{2} - 12 x + 8 y + 47 = 0

(x - 1)^{2} + (y - 2)^{2} = (22)^{2}

x^{2} - 6 x + y^{2} - 4 y = - 9

What is x^2+y^2+4x+8y-36=0 ?
The solution to x^2+y^2+4x+8y-36=0 is Circle with (a,b)=(-2,-4),r=2sqrt(14)