What is x^2+4y+y^2-12=0 ?

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

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x^2+4y+y^2-12=0

x^{2} + 4 y + y^{2} - 12 = 0

(a, b) = (0, - 2), r = 4

Solution steps

x^{2} + 4 y + y^{2} - 12 = 0

Rewrite

x^{2} + 4 y + y^{2} - 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, - 2), r = 4

(x - 5)^{(2)} + (y + 10)^{(2)} = 5^{(2)}

(x - 4)^{2} + (y - (- 3))^{2} = 5^{2}

(x - 1)^{2} + (y - 3)^{2} = 64

2 x^{2} + 2 y^{2} + 12 y + 10 = 0

(x - 7)^{(2)} + (y + 15)^{(2)} = 200

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