What is the center (x-3)^2+(y+3)^2=8 ?

The center (x-3)^2+(y+3)^2=8 is (3,-3)

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center (x-3)^2+(y+3)^2=8

center

(x - 3)^{2} + (y + 3)^{2} = 8

(3, - 3)

Solution steps

(x - 3)^{2} + (y + 3)^{2} = 8

Rewrite

(x - 3)^{2} + (y + 3)^{2} = 8

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 3))^{2} = (8)^{2}

Therefore the circle properties are:

(a, b) = (3, - 3), r = 8

And the center is:

(3, - 3)

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

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

x^{2} + y^{2} - 10 x - 12 y + 45 = 0

x^{2} + y^{2} - 8 x + 2 y + 13 = 0

(x + 4)^{2} + (y - (- 1))^{2} = 12