What is the center x^2+(y+5)^2=100 ?

The center x^2+(y+5)^2=100 is (0,-5)

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center x^2+(y+5)^2=100

center

x^{2} + (y + 5)^{2} = 100

(0, - 5)

Solution steps

x^{2} + (y + 5)^{2} = 100

Rewrite

x^{2} + (y + 5)^{2} = 100

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 5))^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (0, - 5), r = 10

And the center is:

(0, - 5)

(x + 5)^{2} + y^{2} = 64

\frac{x ^{2}}{49} + \frac{y ^{2}}{49} = 1

(x + 2)^{2} + (y - 2)^{2} = 20

4 x^{2} - 40 x + 4 y^{2} - 32 y = 5

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