How can we determine the point at which the distance between vectors is equal to a certain constant? Consider the following points:
$$A(-3,0)\hspace{1cm}
B(3,0)\hspace{1cm}
C(x,y)$$
Now consider the following vectors:
$$CA\hspace{1cm}
CB\hspace{1cm}
CO$$
where $O$ is the origin $O(0,0)$.
Consider the vector $HF$, of magnitude $4$. It is perpendicular to the position vector $OC$ of $C$. All this is depicted in the diagram below. How can the position of $E$ or of $H$ and $F$ be determined with respect to $C(x,y)$ while maintaining a magnitude of $4$? 
TL;DR? Essentially, considering the diagram below and that $C$ is an arbitrary point, where must $E$ be in order to achieve a length of $4$ for $FH$.

Badly worded question, I know, but I can't seem to simplify it nor answer it.
Thanks,
Yazan

I won't do all the calculations since I didn't find a nice method. However, the reasoning I follow is simple to understand, despite involving horrible expressions.
In the sequel, I denote by $d_{XY}$ a line passing thgrouh distincts points $X$ and $Y$. I also take $C(\alpha,\beta)$ to avoid confusion and since we don't look for $C$ but for $E$, so the coordinates of $C$ are parameters.
First case: $C$ does not lie on an axis
As the title says, we consider that $C$ does not lie on an axis. Thus, $\alpha\neq 0$ and $\beta\neq 0$. We can directly deduce the following equations: 
\begin{align*}
d_{CA}\equiv y&=\frac{\beta}{\alpha+3}(x+3)\\
d_{CB}\equiv y&=\frac{\beta}{\alpha-3}(x-3)\\
d_{CO}\equiv y&=\frac{\beta}{\alpha}x
\end{align*}
Note: you have to discuss $\alpha=\pm 3$.
Let $\{d_{p}\vert p\in\Bbb R\}$ be the family of lines that are perpendicular to $d_{CO}$. We have (for any $p\in\Bbb R$):
$$d_{p}\equiv y=\frac{-\alpha}{\beta}x+p$$
since two lines are perpendicular iff the product of their respective slopes is equal to $-1$. Note that there is only one $p$ that will give the "correct" line. We shall be able to express the coordinates of $F$ and $H$ in terms of $p$ and determine $p$ via the condition $\vert HF\vert=4$.
You can now see that $F=d_{CA}\cap d_{p}$ and $H=d_{CB}\cap d_{p}$ for a particular $p$. We have for $d_{CA}\cap d_{p}$:
\begin{align*}
\frac{-\alpha}{\beta}x_{F} +p&=\frac{\beta}{\alpha+3}(x+3)\\
\iff x_{F} &= \frac{p-\frac{3\beta}{\alpha+3}}{\frac{\beta}{\alpha+3}+\frac{\alpha}{\beta}}\\
&=\left(p-\frac{3\beta}{\alpha+3}\right)\frac{\beta(\alpha+3)}{\beta^{2}+\alpha(\alpha+3)}
\end{align*}
Note: you have to discuss the case where the denominators are $0$ separately. This can be partially overcome using parametric equations instead of cartesian ones but I don't know if you know these concepts.
Similarly, for $x_{H}$, we obtain:
$$x_{H}=\left(p+\frac{3\beta}{\alpha-3}\right)\frac{\beta(\alpha-3)}{\beta^{2}+\alpha(\alpha-3)}$$
You then deduce $y_{F}$ and $y_{H}$ using the fact that $F\in d_{CA}$ and $H\in d_{CB}$:
\begin{align*}
y_{F} &= \frac{\beta}{\alpha+3}\left(x_{F}+3\right)\\
y_{H} &= \frac{\beta}{\alpha-3}\left(x_{H}-3\right)
\end{align*}
Now, use the fact that $\vert HF\vert = 4$, that is:
$$(x_{F}-x_{H})^{2}+(y_{F}-y_{H})^{2}=16$$
This equation only depends on $p$ and you can thus determine the latter. I didn't do the calculations but you shall probably have two values for $p$. One of these two values correspond to the central symmetry around $C$. As $E$ will always be closer to the origin than $C$, you can rule one of these two values according to that criterion.
Now that you have determined $p$, you have determined the coordinates of $F$ and $H$ (of course, they depend on $\alpha,\beta$). But you can also directly get $E$ by calculating $d_{p}\cap d_{CO}$. We obtain:
$$E=\left(p\frac{\alpha\beta}{\alpha^{2}+\beta^{2}}\,,\,p\frac{\beta^{2}}{\alpha^{2}+\beta^{2}}\right)$$
Second case: $C$ lies on an axis
We suppose $C\neq O$ (if it is the case, $E=O$). If $C$ lies on the $x$ axis, there is obviously an infinity of solutions that you can easily write down (but $E$ will be on the $x$ axis as well). If $C$ lies on the $y$ axis, that is $C=(0,\beta)$ with $\beta\neq 0$, then $E$ as well. In that case, by the Thales theorem, we know that
$$\frac{\vert FH\vert}{\vert AB\vert}=\frac{4}{6}=\frac{\vert CE\vert}{\vert CO\vert}\iff \vert CE\vert=\frac{2}{3}\vert\beta\vert$$
Now, this yields:
$$y_{E}=\beta\mp\frac{2}{3}\beta$$
We can rule out one of these two values with the fact that $E$ must be closer to the origin than $C$. Thus,
$$y_{E}=\frac{1}{3}\beta$$
