What is the center x^2+y^2+8x-12y+27=0 ?

The center x^2+y^2+8x-12y+27=0 is (-4,6)

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center x^2+y^2+8x-12y+27=0

center

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

(- 4, 6)

Solution steps

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

Rewrite

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - 6)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 4, 6), r = 5

And the center is:

(- 4, 6)

x^{2} + y^{2} + 8 y = 33

x^{2} + y^{2} - 4 x + 2 y - 11 = 0

x^{2} + y^{2} + 6 x - 2 y + 1 = 0

(x - 3)^{2} + (y + 2)^{2} = 10

2 (x - 3)^{2} + 2 y^{2} = 8

What is the center x^2+y^2+8x-12y+27=0 ?
The center x^2+y^2+8x-12y+27=0 is (-4,6)