What is the center (x-3)^2+(y+2)^2=5 ?

The center (x-3)^2+(y+2)^2=5 is (3,-2)

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center (x-3)^2+(y+2)^2=5

center

(x - 3)^{2} + (y + 2)^{2} = 5

(3, - 2)

Solution steps

(x - 3)^{2} + (y + 2)^{2} = 5

Rewrite

(x - 3)^{2} + (y + 2)^{2} = 5

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 2))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (3, - 2), r = 5

And the center is:

(3, - 2)

(x + 6)^{2} + (y - 1)^{2} = 16

(x + 6)^{2} + (y - 7)^{2} = 81

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

(x - 1.5)^{2} + (y + 0.5)^{2} = 110.5

(x - 7)^{2} + (y + 4)^{2} = 9