What is the center x^2+y^2+14x+16y+14=0 ?

The center x^2+y^2+14x+16y+14=0 is (-7,-8)

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center x^2+y^2+14x+16y+14=0

center

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

(- 7, - 8)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 7))^{2} + (y - (- 8))^{2} = (311)^{2}

Therefore the circle properties are:

(a, b) = (- 7, - 8), r = 311

And the center is:

(- 7, - 8)

5 x^{2} + 5 y^{2} + 10 x - 5 y + 3 = 0

(x - 126)^{2} + (y - 102.5)^{2} \geq 149. 5^{2}

(x + 2)^{2} + (y - 5)^{2} = 7

x (x - 12) + y (y + 2) = 12

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

What is the center x^2+y^2+14x+16y+14=0 ?
The center x^2+y^2+14x+16y+14=0 is (-7,-8)