What is the center x^2+y^2+14x-2y+25=0 ?

The center x^2+y^2+14x-2y+25=0 is (-7,1)

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center x^2+y^2+14x-2y+25=0

center

x^{2} + y^{2} + 14 x - 2 y + 25 = 0

(- 7, 1)

Solution steps

x^{2} + y^{2} + 14 x - 2 y + 25 = 0

Rewrite

x^{2} + y^{2} + 14 x - 2 y + 25 = 0

in the form of the standard circle equation

(x - (- 7))^{2} + (y - 1)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 7, 1), r = 5

And the center is:

(- 7, 1)

(x - 1)^{2} + (y - 4)^{2} = 25

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

(x + 3)^{2} + (y - 4)^{2} = 7

x^{2} + y^{2} - 10 x + 6 y - 2 = 0

x^{2} + y^{2} = 3 x + 4 y

What is the center x^2+y^2+14x-2y+25=0 ?
The center x^2+y^2+14x-2y+25=0 is (-7,1)