Let $O,A$ be two points in the plane with $|\vec{OA}|=3$. Which line is formed by the points $M$ of the plane, such that $\vec{OM}(\vec{OM}-2\vec{OA})=7$ ?
My attempt..
Suppose $\vec{OM}=\vec{x}$ and $\vec{OA}=\vec{y}$. Then the above equation can be rewritten as $C: \vec{x}^2-2\vec{x}\vec{y}-7=0$. The discriminant is $64$, thus $\vec{x}=\vec{y}\pm 4\implies \vec{OM}=\vec{OA}\pm 4$. So for every value of $\vec{OA}$ there are two values of $\vec{OM}$..
Here's where i'm stuck. I know that $C$ is a parabola, so the points $M$ must lie in a parabola?? I want to understand what i have to do here and in general i want to know if what i did makes sense.. What have i accomplished really? I'm confused. Please help. Thanks in advance.
I found the answer. Here goes..
By letting $A$ be the point of reference we have:
$\vec{OM}(\vec{OM}-2\vec{OA})=7 \implies (\vec{AM}-\vec{AO})(\vec{AM}-\vec{AO}-2\vec{OA})=7 \implies (\vec{AM}-\vec{AO})(\vec{AM}+\vec{AO})=7 \implies |\vec{AM}|^2 -|\vec{AO}|^2=7 \implies |\vec{AM}|^2=7+9=16 \implies |\vec{AM}|=4$
Therefore we notice that the points $M$ are located $4$ units around $A$, which means they are in a circle with center $A$ and radius $4$.