What is the center x^2+y^2+14x+12y+21=0 ?

The center x^2+y^2+14x+12y+21=0 is (-7,-6)

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center x^2+y^2+14x+12y+21=0

center

x^{2} + y^{2} + 14 x + 12 y + 21 = 0

(- 7, - 6)

Solution steps

x^{2} + y^{2} + 14 x + 12 y + 21 = 0

Rewrite

x^{2} + y^{2} + 14 x + 12 y + 21 = 0

in the form of the standard circle equation

(x - (- 7))^{2} + (y - (- 6))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (- 7, - 6), r = 8

And the center is:

(- 7, - 6)

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

(x + 4)^{2} + (y - \frac{3}{4})^{2} = 1

(x - 2)^{2} + (y + 1)^{2} = 16

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

What is the center x^2+y^2+14x+12y+21=0 ?
The center x^2+y^2+14x+12y+21=0 is (-7,-6)