What is the center (x-3)^2+(y+9)^2=9^2 ?

The center (x-3)^2+(y+9)^2=9^2 is (3,-9)

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center (x-3)^2+(y+9)^2=9^2

center

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

(3, - 9)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 9), r = 9

And the center is:

(3, - 9)

x^{2} + y^{2} = 289

(x + 1)^{2} + (y + 2)^{2} = 25

x^{2} + y^{2} - 20 x + 64 y - 371 = 0

4 x^{2} + 4 y^{2} - 8 y - 5 = 0

(x - 1)^{2} + (y - 4)^{2} = 9

What is the center (x-3)^2+(y+9)^2=9^2 ?
The center (x-3)^2+(y+9)^2=9^2 is (3,-9)