What is the center x^2+y^2+12x-14y+84=0 ?

The center x^2+y^2+12x-14y+84=0 is (-6,7)

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center x^2+y^2+12x-14y+84=0

center

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

(- 6, 7)

Solution steps

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

Rewrite

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - 7)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 6, 7), r = 1

And the center is:

(- 6, 7)

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}

x^{2} + y^{2} + 14 x - 16 + 64 = 0

x^{2} + 8 x + y^{2} + 4 y = 4

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

What is the center x^2+y^2+12x-14y+84=0 ?
The center x^2+y^2+12x-14y+84=0 is (-6,7)