What is the center x^2+y^2+4x-6y-12=0 ?

The center x^2+y^2+4x-6y-12=0 is (-2,3)

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center x^2+y^2+4x-6y-12=0

center

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

(- 2, 3)

Solution steps

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

Rewrite

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 5

And the center is:

(- 2, 3)

x^{2} + y^{2} - 4 x - 8 y + 18 = 0

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

(x + 5)^{2} + (y - 2)^{2} = 25

x^{2} + y^{2} + 4 x + 6 y = 23

(x - 5)^{2} + (y + 2)^{2} = 15