What is (x-2)^2+(y-4)^2=16 ?

The solution to (x-2)^2+(y-4)^2=16 is Circle with (a,b)=(2,4),r=4

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(x-2)^2+(y-4)^2=16

(x - 2)^{2} + (y - 4)^{2} = 16

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

Solution steps

(x - 2)^{2} + (y - 4)^{2} = 16

Rewrite

(x - 2)^{2} + (y - 4)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x + 4)^{2} + (y + 1)^{2} = 25

x^{2} + y^{2} - 8 x + 14 y + 40 = 0

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

3 x^{2} + 3 y^{2} - 18 x + 12 y - 36 = 0

(x - 4)^{2} + (y + 1)^{2} = 25

What is (x-2)^2+(y-4)^2=16 ?
The solution to (x-2)^2+(y-4)^2=16 is Circle with (a,b)=(2,4),r=4