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

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

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

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

(a, b) = (2, - 4), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 4), r = 2

x^{2} + y^{2} \geq 5

x^{2} + y^{2} - 6 x + 6 y + 14 = 0

(x - 3)^{2} + (y^{2} - 2^{2}) = 0

(x - 4)^{2} + y^{2} = 36

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

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