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

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

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

center

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

(5, - 7)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(5, - 7)

3 x^{2} + 3 y^{2} - 12 x - 2 = 0

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35

3 x^{2} + 3 y^{2} - 18 x - 30 y - 6 = 0

(x - 4)^{2} + (y - 1)^{2} = 218