What is x^2+y^2+4y+3=0 ?

The solution to x^2+y^2+4y+3=0 is Circle with (a,b)=(0,-2),r=1

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x^2+y^2+4y+3=0

x^{2} + y^{2} + 4 y + 3 = 0

(a, b) = (0, - 2), r = 1

Solution steps

x^{2} + y^{2} + 4 y + 3 = 0

Rewrite

x^{2} + y^{2} + 4 y + 3 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, - 2), r = 1

(x + 4)^{2} + (y - 2)^{2} = 16

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

x^{2} + y^{2} - 10 x + 6 y - 15 = 0

4 (x - 5)^{2} + 4 y^{2} = 64

3 x^{2} - 6 x + 3 y^{2} - 18 y - 78 = 0

What is x^2+y^2+4y+3=0 ?
The solution to x^2+y^2+4y+3=0 is Circle with (a,b)=(0,-2),r=1