What is the center x^2+y^2+6y+2=0 ?

The center x^2+y^2+6y+2=0 is (0,-3)

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center x^2+y^2+6y+2=0

center

x^{2} + y^{2} + 6 y + 2 = 0

(0, - 3)

Solution steps

x^{2} + y^{2} + 6 y + 2 = 0

Rewrite

x^{2} + y^{2} + 6 y + 2 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 3))^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (0, - 3), r = 7

And the center is:

(0, - 3)

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

x^{2} + y^{2} - 6 x + 12 y + 9 = 9

x^{2} + 7 x + y^{2} - 10 y + 6 = 0

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

(x - 5)^{2} + (y - 3)^{2} = 64