What is the center (x-5)^2+(y+1)^2=12 ?

The center (x-5)^2+(y+1)^2=12 is (5,-1)

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center (x-5)^2+(y+1)^2=12

center

(x - 5)^{2} + (y + 1)^{2} = 12

(5, - 1)

Solution steps

(x - 5)^{2} + (y + 1)^{2} = 12

Rewrite

(x - 5)^{2} + (y + 1)^{2} = 12

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 1))^{2} = (12)^{2}

Therefore the circle properties are:

(a, b) = (5, - 1), r = 12

And the center is:

(5, - 1)

x^{2} + y^{2} + 6 x - 16 y + 48 = 0

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

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

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

(x - 0)^{2} + (y - 0)^{2} = 4