What is the center (x-3)^2+(y+1)^2=25 ?

The center (x-3)^2+(y+1)^2=25 is (3,-1)

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center (x-3)^2+(y+1)^2=25

center

(x - 3)^{2} + (y + 1)^{2} = 25

(3, - 1)

Solution steps

(x - 3)^{2} + (y + 1)^{2} = 25

Rewrite

(x - 3)^{2} + (y + 1)^{2} = 25

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

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

x^{2} + y^{2} = 8 (x + y) + 4

(\frac{x - 7}{4})^{2} + (\frac{y + 3}{4})^{2} = \frac{169}{16}

x^{2} + y^{2} + 4 x + 6 y + 9 = 0