What is the center (x-6)^2+(y-4)^2=5^2 ?

The center (x-6)^2+(y-4)^2=5^2 is (6,4)

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center (x-6)^2+(y-4)^2=5^2

center

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

(6, 4)

Solution steps

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

Rewrite

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

in the form of the standard circle equation

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (6, 4), r = 5

And the center is:

(6, 4)

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

x^{2} + 8 x + y^{2} + 4 y - 4 = 0

4 x^{2} + 40 x + 4 y^{2} = 0

(x - 7)^{2} + (y - 5)^{2} = 4

x^{2} + y^{2} - 10 x + 16 y + 85 = 0

What is the center (x-6)^2+(y-4)^2=5^2 ?
The center (x-6)^2+(y-4)^2=5^2 is (6,4)