What is the center x^2+y^2-14x-12y+59=0 ?

The center x^2+y^2-14x-12y+59=0 is (7,6)

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center x^2+y^2-14x-12y+59=0

center

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

(7, 6)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 7)^{2} + (y - 6)^{2} = (26)^{2}

Therefore the circle properties are:

(a, b) = (7, 6), r = 26

And the center is:

(7, 6)

(x + 5)^{2} + (y + 1)^{2} = 9

x^{2} + y^{2} - 49 = 0

(x + 1)^{2} + (y - 2)^{2} = 9

x^{2} + y^{2} + 4 x - 2 y - 31 = 0

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

What is the center x^2+y^2-14x-12y+59=0 ?
The center x^2+y^2-14x-12y+59=0 is (7,6)