What is the center (x-5)^2+(y-3)^2=64 ?

The center (x-5)^2+(y-3)^2=64 is (5,3)

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center (x-5)^2+(y-3)^2=64

center

(x - 5)^{2} + (y - 3)^{2} = 64

(5, 3)

Solution steps

(x - 5)^{2} + (y - 3)^{2} = 64

Rewrite

(x - 5)^{2} + (y - 3)^{2} = 64

in the form of the standard circle equation

(x - 5)^{2} + (y - 3)^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (5, 3), r = 8

And the center is:

(5, 3)

12 x^{2} + 12 y^{2} - 18 x + 4 y - 5 = 0

(x - 3)^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} - 10 x + 9 y + 1 x = 0

x^{2} - 7 x + y^{2} + 8 y - 14 = 0

3 (x - 5)^{2} + 3 (y + 9)^{2} = 48