What is the difference between the equations $x^2=4y^2$ and $x^2=2yx$ Suppose I am given the equation $x=2y$.
My interpretation of this equation is $A=\{\langle x,y \rangle \in \mathbb R \times \mathbb R |\  x=2y\ \}$
Given this interpretation, I want to know the difference between the equations $x^2=4y^2$ and $x^2=2yx$.

For the equation $x^2=2yx$, we have a single line that completely overlaps with $x=2y$.
For the equation $x^2=4y^2$, we have two intersecting lines, only one of which overlaps with $x=2y$.
In either case, this is satisfying because it explains the logical concepts of:
$$x=2y \implies x^2=4y^2$$ $$\text{and}$$ $$x=2y \implies x^2=2yx$$ which I feel as though I am implicitly using during different algebraic proofs. That is to say, the set of solutions for $x=2y$ also satisfies (intersects with) the equations (sets) $x^2=4y^2$ and $x^2=2yx$.
In line with our earlier set interpretation, letting $x^2=2yx$ represent the set $B=\{\langle x,y \rangle \in \mathbb R \times \mathbb R | x^2=2yx\}$ and letting $x^2=4y^2$ represent the set $C=\{\langle x,y \rangle \in \mathbb R \times \mathbb R | x^2=4y^2\}$, the aforementioned implication suggests $A \subseteq B$ and $A\subseteq C$.
However, given my descriptions of the graphs, it is quite clear that we can be more specific...namely:
$A=B$ and $A \subsetneqq C$

I see that for the equation $x^2=4y^2$, square rooting both sides yields $\pm x=2y$, which explains its graph of the two intersecting lines. But I am wondering if there is a more fundamental reason that explains this observation. In particular, the quality of inverse exponents versus the quality of inverse multiplication.
To simplify $x^2=2yx$, I am effectively multiplying each side by $x^{-1}$. Comparatively, to simplify $x^2=4y^2$ (perhaps better written as $x^2=(2y)^2$), I am taking the inverse operation of squaring. It seems like the inverse operation of squaring yields "more solutions" than the inverse operation of multiplication.
Any added insight would be greatly appreciated!
 A: It is easier to compare if you set $0$ alone on one side.
Your first equation is $x-2y=0$. The two next equations are $x^2-4y^2=0$ and $x^2-2xy=0$. So what to do with those equations? Factor them!
We have $x^2-4y^2=(x-2y)(x+2y)=0$. Since a product is zero iff one of the factors is zero, this equation describes the union of two lines: $x-2y=0$ and $x+2y=0$.
We have $x^2-2xy=x(x-2y)=0$. Since a product is zero iff one of the factors is zero, this equation describes the union of two lines: $x-2y=0$ and $x=0$.
At least to me, things become a lot clearer if we have $0$ on one side of the equation, and factor. These two actions work very nice together in order to couple what's going on with the algebra with what's going on with the geometry. It is not always this easy to factor, but the principle remains.
A: $\displaystyle x^2=4y^2$
$\iff$ $\;y=\pm\frac12x$


$\displaystyle x^2=2yx$
$\iff$ $\;x=0\quad$ or $\quad y=\frac12x$

A: Well if $x= 2y$ then $x^2 =4y^2$ but $x^2=4y^2 \not \implies x = 2y$. (But it does imply $x = \pm 2y$).
So $A = \{(x,y)| x=2y\}\subsetneq \{(x,y)|x^2 =4y^2\}=B$.  If we want to interpret these sets $A $ is a line and $B$ is two crossed lines; one of them is $A$ and the other is the mirror image of $A$.
And also $x=2y \implies x^2 = 2xy$ but $x^2 =2xy \not \implies x=2y$. (But it implies: If $x\ne 0$ then $x = 2y$. But if $x = 0$ then $y$ may be any value.)
So again $A = =\{(x,y)|x=2y\}\subsetneq \{(x,y)|x^2 =2xy\}=C$.  If we want to interpret these sets $A$ is a line and $B$ is two crossed lines: one of them is $A$ and the other is the vertical line $x=0$.
As $A \subset B$ and $A \subset C$ then $A\subset B\cap C$.  And in this case we have
If $x^2 = 4y^2=2xy$ then we have $x^2 = 4y^2 \implies x =\pm 2y$ and $x^2=2xy\implies x=0$ or $x = 2y$ we have $x=2y$ or $x=0=-2y$.  So $x = 2y$.
So $B\cap C = A$ in this case.
....
SO that about covers is.
$x^2 =4y^2$ includes all of $x=2y$ but also adds $x = -2y$ solutions.  $x^2 = 2xy$ includes all of $x = 2y$ but also adds $x =0;y=anything$.
So $x^2 = 4y^2$ and $x^2 = 2xy$ have $x =2y$ in common and $x=0=-2y$ in common. But $x=0=-2y$ is part of $x =2y$ so that only have $x=2y$ in common.
A: Suppose
$$x = 2y \tag{1}\label{eq1}$$
Then certainly
$$x^2 = 4y^2 \tag{2}\label{eq2}$$
Now, if all we know is $x^2 = 4y^2$, then we would know either $x = 2y$ or $x = -2y$. However, we know more than this. We can eliminate $x = -2y$ because of $\eqref{eq1}$.
$x = 2y$ doesn't stop being true even after squaring both sides. \eqref{eq2} implies two posibilies, one of which we can eliminate because of \eqref{eq1}.
(To be more precise, $x=-2y$ is eliminated for $x, y \neq 0$. The solution $x = y = 0$ is fine, and is already covered by $x = 2y$.)
