What is the center (x+3)^2+(y-1)^2=5 ?

The center (x+3)^2+(y-1)^2=5 is (-3,1)

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center (x+3)^2+(y-1)^2=5

center

(x + 3)^{2} + (y - 1)^{2} = 5

(- 3, 1)

Solution steps

(x + 3)^{2} + (y - 1)^{2} = 5

Rewrite

(x + 3)^{2} + (y - 1)^{2} = 5

in the form of the standard circle equation

(x - (- 3))^{2} + (y - 1)^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (- 3, 1), r = 5

And the center is:

(- 3, 1)

x^{2} - 10 x + y^{2} + 8 y = - 20

2 x^{2} + 2 y^{2} - 4 x + 12 y - 24 = 0

x^{2} + 12 x + y^{2} + 12 y + 71 = 0

x^{2} + 6 x + 1 + y^{2} = 4

(x + 4)^{2} + (y + 2)^{2} = 25