What is the center x^2+y^2-8x+12y=-43 ?

The center x^2+y^2-8x+12y=-43 is (4,-6)

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center x^2+y^2-8x+12y=-43

center

x^{2} + y^{2} - 8 x + 12 y = - 43

(4, - 6)

Solution steps

x^{2} + y^{2} - 8 x + 12 y = - 43

Rewrite

x^{2} + y^{2} - 8 x + 12 y = - 43

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, - 6), r = 3

And the center is:

(4, - 6)

16 x^{2} + 16 y^{2} + 96 x - 40 y = 315

x^{2} + y^{2} - 2 x + 6 y + 9 = 0

x^{2} + y^{2} - 4 x + 10 y + 13 = 0

x^{2} + y^{2} + 6 x - 2 y - 15 = 0

8 x + x^{2} - 2 y = 64 - y^{2}