What is the center x^2+y^2+6x-4y-3=0 ?

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

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center x^2+y^2+6x-4y-3=0

center

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

(- 3, 2)

Solution steps

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

Rewrite

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 2)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 3, 2), r = 4

And the center is:

(- 3, 2)

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

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

x^{2} + y^{2} - 12 y + 32 = 0

x^{2} + y^{2} - 25 = 0

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