What is x^2+y^2+8y-128=0 ?

The solution to x^2+y^2+8y-128=0 is Circle with (a,b)=(0,-4),r=12

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x^2+y^2+8y-128=0

x^{2} + y^{2} + 8 y - 128 = 0

(a, b) = (0, - 4), r = 12

Solution steps

x^{2} + y^{2} + 8 y - 128 = 0

Rewrite

x^{2} + y^{2} + 8 y - 128 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 4))^{2} = 1 2^{2}

Therefore the circle properties are:

(a, b) = (0, - 4), r = 12

2 x^{2} + 2 y^{2} + 8 x - 12 y - 6 = 0

(x - 6)^{2} + (y - 9)^{2} = 49

x^{2} + y^{2} + 12 x - 2 y - 1 = 0

(x + 2)^{2} + (y + 1)^{2} = 25

x^{2} + (y + 1)^{2} = 7

What is x^2+y^2+8y-128=0 ?
The solution to x^2+y^2+8y-128=0 is Circle with (a,b)=(0,-4),r=12