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

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

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

center

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

(- 2, 3)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 2, 3)

9 x^{2} + 9 y^{2} + 18 x - 18 y + 14 = 0

(x + 3)^{2} - 49 = - y^{2}

y^{2} = 4 - x^{2}

(x + \frac{5}{2})^{2} + (y - 14)^{2} = 20

x^{2} + y^{2} = 100