What is the center (x+6)^2+(y+14)^2=400 ?

The center (x+6)^2+(y+14)^2=400 is (-6,-14)

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center (x+6)^2+(y+14)^2=400

center

(x + 6)^{2} + (y + 14)^{2} = 400

(- 6, - 14)

Solution steps

(x + 6)^{2} + (y + 14)^{2} = 400

Rewrite

(x + 6)^{2} + (y + 14)^{2} = 400

in the form of the standard circle equation

(x - (- 6))^{2} + (y - (- 14))^{2} = 2 0^{2}

Therefore the circle properties are:

(a, b) = (- 6, - 14), r = 20

And the center is:

(- 6, - 14)

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

x^{2} + y^{2} + 2 x + 18 y - 143 = 0

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

x^{2} + y^{2} - 2 x - 15 = 0

4 x^{2} + 4 y^{2} - 16 x + 24 y - 12 = 0

What is the center (x+6)^2+(y+14)^2=400 ?
The center (x+6)^2+(y+14)^2=400 is (-6,-14)