What is the center x^2+y^2+14x-12y+4=0 ?

The center x^2+y^2+14x-12y+4=0 is (-7,6)

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center x^2+y^2+14x-12y+4=0

center

x^{2} + y^{2} + 14 x - 12 y + 4 = 0

(- 7, 6)

Solution steps

x^{2} + y^{2} + 14 x - 12 y + 4 = 0

Rewrite

x^{2} + y^{2} + 14 x - 12 y + 4 = 0

in the form of the standard circle equation

(x - (- 7))^{2} + (y - 6)^{2} = 9^{2}

Therefore the circle properties are:

(a, b) = (- 7, 6), r = 9

And the center is:

(- 7, 6)

(x + 2)^{2} + (y - 4)^{2} = 25

x^{2} + y^{2} + 2 x - 4 y = 76

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

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

x^{2} + y^{2} + 4 x + 4 y + 7 = 0

What is the center x^2+y^2+14x-12y+4=0 ?
The center x^2+y^2+14x-12y+4=0 is (-7,6)