What is the center 364+28y+y^2+x^2=-26x ?

The center 364+28y+y^2+x^2=-26x is (-13,-14)

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center 364+28y+y^2+x^2=-26x

center

364 + 28 y + y^{2} + x^{2} = - 26 x

(- 13, - 14)

Solution steps

364 + 28 y + y^{2} + x^{2} = - 26 x

Rewrite

364 + 28 y + y^{2} + x^{2} = - 26 x

in the form of the standard circle equation

(x - (- 13))^{2} + (y - (- 14))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 13, - 14), r = 1

And the center is:

(- 13, - 14)

x^{2} + y^{2} = 12

x^{2} + y^{2} = 11

(x + 4)^{2} + (y - 6)^{2} = 18

x^{2} + y^{2} = 20

x^{2} + y^{2} + 8 x = 0

What is the center 364+28y+y^2+x^2=-26x ?
The center 364+28y+y^2+x^2=-26x is (-13,-14)