What is the center x^2+(y+20)^2=100 ?

The center x^2+(y+20)^2=100 is (0,-20)

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center x^2+(y+20)^2=100

center

x^{2} + (y + 20)^{2} = 100

(0, - 20)

Solution steps

x^{2} + (y + 20)^{2} = 100

Rewrite

x^{2} + (y + 20)^{2} = 100

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 20))^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (0, - 20), r = 10

And the center is:

(0, - 20)

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

(x + 3)^{2} + (y + 5)^{2} = 4

(x + 3)^{2} + (y + 5)^{2} = 9

x^{2} + y^{2} - 10 x + 9 y + 1 = 0

(x - 1)^{2} + (y + 3)^{2} = 9