What is (x-2)^{(2)}+(y+1)^{(2)}>= 4 ?

The solution to (x-2)^{(2)}+(y+1)^{(2)}>= 4 is Circle with (a,b)=(2,-1),r=2

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(x-2)^{(2)}+(y+1)^{(2)}>= 4

(x - 2)^{(2)} + (y + 1)^{(2)} \geq 4

(a, b) = (2, - 1), r = 2

Solution steps

(x - 2)^{2} + (y + 1)^{2} \geq 4

Rewrite

(x - 2)^{2} + (y + 1)^{2} \geq 4

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 1), r = 2

(x - 2)^{2} + (y - 6)^{2} = 16

(x - 3)^{2} + (y - 5)^{2} = 6^{2}

(x + \frac{54}{13})^{2} + (y - \frac{123}{26})^{2} = \frac{5165}{52}

(x + 3)^{2} + y^{2} = 48

(x - 4)^{2} + (y + 5)^{2} - 31 = 0

What is (x-2)^{(2)}+(y+1)^{(2)}>= 4 ?
The solution to (x-2)^{(2)}+(y+1)^{(2)}>= 4 is Circle with (a,b)=(2,-1),r=2