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

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

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

center

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

(1, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 2), r = 3

And the center is:

(1, - 2)

x^{2} + z^{2} + 4 x - 4 z = - 4

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

(x - 1)^{2} + (y - 5)^{2} = 17

(x - 2)^{2} + (y + 3)^{2} = \frac{16}{9}

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