What is the center (x-126)^2+(y-102.5)^2>= 149.5^2 ?

The center (x-126)^2+(y-102.5)^2>= 149.5^2 is (126,102.5)

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center (x-126)^2+(y-102.5)^2>= 149.5^2

center

(x - 126)^{2} + (y - 102.5)^{2} \geq 149. 5^{2}

(126, 102.5)

Solution steps

(x - 126)^{2} + (y - 102.5)^{2} \geq 149. 5^{2}

Rewrite

(x - 126)^{2} + (y - 102.5)^{2} \geq 149. 5^{2}

in the form of the standard circle equation

(x - 126)^{2} + (y - 102.5)^{2} = 149. 5^{2}

Therefore the circle properties are:

(a, b) = (126, 102.5), r = 149.5

And the center is:

(126, 102.5)

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

x (x - 12) + y (y + 2) = 12

(x - 6)^{2} + (y - 4)^{2} = 5^{2}

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

x^{2} + 8 x + y^{2} + 4 y - 4 = 0

What is the center (x-126)^2+(y-102.5)^2>= 149.5^2 ?
The center (x-126)^2+(y-102.5)^2>= 149.5^2 is (126,102.5)