What is the center x^2+y^2+4x+6y+9=0 ?

The center x^2+y^2+4x+6y+9=0 is (-2,-3)

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center x^2+y^2+4x+6y+9=0

center

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

(- 2, - 3)

Solution steps

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

Rewrite

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, - 3), r = 2

And the center is:

(- 2, - 3)

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0

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

x^{2} + y^{2} + 2 x + 10 y + 18 = 0

x^{2} + (y - 1)^{2} = 9

(x - 11)^{2} + (y + 2)^{2} = 16