What is the center x^2+y^2+4x-6y-23=0 ?

The center x^2+y^2+4x-6y-23=0 is (-2,3)

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center x^2+y^2+4x-6y-23=0

center

x^{2} + y^{2} + 4 x - 6 y - 23 = 0

(- 2, 3)

Solution steps

x^{2} + y^{2} + 4 x - 6 y - 23 = 0

Rewrite

x^{2} + y^{2} + 4 x - 6 y - 23 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 3)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- 2, 3), r = 6

And the center is:

(- 2, 3)

2 x^{2} + 2 y^{2} - 8 x + 7 = 0

4 x^{2} + 4 y^{2} - 16 x + 40 y + 76 = 0

x^{2} + y^{2} \leq 9

x^{2} + 10 x + y^{2} - 16 - 11 = 0

x^{2} + y^{2} + 10 x - 8 y + 25 = 0