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

The solution to x^{(2)}+(y-4)^{(2)}=16 is Circle with (a,b)=(0,4),r=4

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

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

(a, b) = (0, 4), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 4), r = 4

x^{2} + y^{2} + 5 = 6 y

x^{2} + y^{2} - 60 x + 16 y + 739 = 0

- y^{2} - x^{2} = - 9

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

2 x^{2} + 2 y^{2} - 16 y + 24 = 0

What is x^{(2)}+(y-4)^{(2)}=16 ?
The solution to x^{(2)}+(y-4)^{(2)}=16 is Circle with (a,b)=(0,4),r=4