What is the center (x-3)^2+(y+5)^2=49 ?

The center (x-3)^2+(y+5)^2=49 is (3,-5)

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center (x-3)^2+(y+5)^2=49

center

(x - 3)^{2} + (y + 5)^{2} = 49

(3, - 5)

Solution steps

(x - 3)^{2} + (y + 5)^{2} = 49

Rewrite

(x - 3)^{2} + (y + 5)^{2} = 49

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 5))^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (3, - 5), r = 7

And the center is:

(3, - 5)

10 x^{2} + 10 y^{2} - 10 x - y - 137 = 0

x^{2} + y^{2} - 10 x + 4 y + 20 = 0

x^{2} + y^{2} + 12 x - 14 y + 84 = 0

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}

x^{2} + y^{2} + 14 x - 16 + 64 = 0