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

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

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

center

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

(- 3, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 3, 2)

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x^{2} + y^{2} + 8 x - 2 y - 19 = 0