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} + y^{2} + 4 x - 2 y - 31 = 0

364 + 28 y + y^{2} + x^{2} = - 26 x

x^{2} + y^{2} = 12

x^{2} + y^{2} = 11

(x + 4)^{2} + (y - 6)^{2} = 18