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

The solution to (2x)^2+(2y-1)^2=4^2 is Circle with (a,b)=(0, 1/2),r=2

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(2x)^2+(2y-1)^2=4^2

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

(a, b) = (0, \frac{1}{2}), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - \frac{1}{2})^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (0, \frac{1}{2}), r = 2

x^{2} + 20 x + y^{2} + 16 = - 20

x^{2} - 8 x + y^{2} + 4 y - 5 = 0

2 x^{2} + 2 y^{2} - 1 = 0

x^{2} - 14 x + y^{2} + 8 y + 53 = 0

x^{2} + y^{2} = 259

What is (2x)^2+(2y-1)^2=4^2 ?
The solution to (2x)^2+(2y-1)^2=4^2 is Circle with (a,b)=(0, 1/2),r=2