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

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

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x^2+y^2+4y=4

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

(a, b) = (0, - 2), r = 22

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 2))^{2} = (22)^{2}

Therefore the circle properties are:

(a, b) = (0, - 2), r = 22

4 x^{2} + 36 x + 4 y^{2} - 8 y + 68 = x^{2} - 6 x + y^{2} - 4 y + 13

x^{2} + (y - 1)^{2} = 16

(x^{2} - 10 x + 25) + (y^{2} + 4 y + 4) = 36

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

x^{2} + y^{2} - 6 x + 12 y + 41 = 0

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