What is the center x^2+y^2-10x-16y+53=0 ?

The center x^2+y^2-10x-16y+53=0 is (5,8)

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center x^2+y^2-10x-16y+53=0

center

x^{2} + y^{2} - 10 x - 16 y + 53 = 0

(5, 8)

Solution steps

x^{2} + y^{2} - 10 x - 16 y + 53 = 0

Rewrite

x^{2} + y^{2} - 10 x - 16 y + 53 = 0

in the form of the standard circle equation

(x - 5)^{2} + (y - 8)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (5, 8), r = 6

And the center is:

(5, 8)

x^{2} + y^{2} - 12 x + 8 y - 5 = 0

(x - 5)^{2} + (y - 3)^{2} = 49

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

(x + 2)^{2} + (y + 3)^{2} = 9

(x + 2)^{2} + (y - 5)^{2} = 4

What is the center x^2+y^2-10x-16y+53=0 ?
The center x^2+y^2-10x-16y+53=0 is (5,8)