What is the center (x-7)^2+(y+5)^2=4 ?

The center (x-7)^2+(y+5)^2=4 is (7,-5)

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center (x-7)^2+(y+5)^2=4

center

(x - 7)^{2} + (y + 5)^{2} = 4

(7, - 5)

Solution steps

(x - 7)^{2} + (y + 5)^{2} = 4

Rewrite

(x - 7)^{2} + (y + 5)^{2} = 4

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 5))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (7, - 5), r = 2

And the center is:

(7, - 5)

(x + 4)^{2} + (y - 2)^{2} = 121

3 (x + \frac{3}{4})^{2} + 3 (y - \frac{1}{2})^{2} = 36

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

(x + 4)^{2} + (y - 1)^{2} = 25

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