if $x \neq 0$ and $x = \sqrt{4xy - 4y^2}$, then how does expressing $x$ in terms of $y$ mean $x = 2y$?

Question

I have the equation $x=\sqrt{4xy - 4y^2}$, and I know that $x=2y$ when expressed in terms of $y$, but I'm not sure of the process to get there.

I know that \begin{align}\sqrt{4xy - 4y^2} &= \sqrt{4y(x-y)}\\ &= 2 \sqrt{y(x-y)} \\ &= 2(xy-y^2)^{\frac12} \end{align}

but pretty stumped.

Answer 1

$$x=\sqrt{4xy - 4y^2}$$ $$x^2=4xy-4y^2$$ $$x^2-4xy+4y^2=0$$ $$(x-2y)^2=0$$ $$x-2y=0$$ $$x=2y$$

Answer 2

Squaring both sides,

$$x^2=4xy-4y^2$$ $$x^2-4xy+4y^2=0$$ $$(x-2y)^2=0$$

Answer 3

$x =\sqrt{4xy + 4y^2}$

$x^2 = 4xy - 4y^2$

$x^2 - 4xy + 4y^2 = 0$

$x = \frac{4y \pm \sqrt{16y^2 - 16y^2}}{2}$

$x = \frac{4y \pm \sqrt{0}}{2}$

$x = 2y$