What is the center (x-1)^2+(y+3)^2=9 ?

The center (x-1)^2+(y+3)^2=9 is (1,-3)

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center (x-1)^2+(y+3)^2=9

center

(x - 1)^{2} + (y + 3)^{2} = 9

(1, - 3)

Solution steps

(x - 1)^{2} + (y + 3)^{2} = 9

Rewrite

(x - 1)^{2} + (y + 3)^{2} = 9

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 3), r = 3

And the center is:

(1, - 3)

x^{2} + y (y + 4) = 10 x + 7

x^{2} + y^{2} - 12 x + 6 y + 36 = 0

9 = 2 y - y^{2} - 6 x - x^{2}

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

(x - 1)^{2} = 27 - (y + 1)^{2}