What is the center (x-4)^2+(y-15)^2=169 ?

The center (x-4)^2+(y-15)^2=169 is (4,15)

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center (x-4)^2+(y-15)^2=169

center

(x - 4)^{2} + (y - 15)^{2} = 169

(4, 15)

Solution steps

(x - 4)^{2} + (y - 15)^{2} = 169

Rewrite

(x - 4)^{2} + (y - 15)^{2} = 169

in the form of the standard circle equation

(x - 4)^{2} + (y - 15)^{2} = 1 3^{2}

Therefore the circle properties are:

(a, b) = (4, 15), r = 13

And the center is:

(4, 15)

(x - 2)^{2} + (y + 5)^{2} = 16

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

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

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

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

What is the center (x-4)^2+(y-15)^2=169 ?
The center (x-4)^2+(y-15)^2=169 is (4,15)