What is the center (x+1)^2+(y-5)^2=4 ?

The center (x+1)^2+(y-5)^2=4 is (-1,5)

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center (x+1)^2+(y-5)^2=4

center

(x + 1)^{2} + (y - 5)^{2} = 4

(- 1, 5)

Solution steps

(x + 1)^{2} + (y - 5)^{2} = 4

Rewrite

(x + 1)^{2} + (y - 5)^{2} = 4

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 5), r = 2

And the center is:

(- 1, 5)

(x + 1)^{2} + (y + 3)^{2} = 9

(x - 5)^{2} + (y + 1)^{2} = 16

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

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

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