If $c_1 = c_2$ the answer is a line $$\frac{O_1+O_2}{2}+t\overline{O_1O_2}$$. I denote $\overline{A}$ and $A$ to be orthogonal vectors.
When $c_1 \ne c_2$ it seems that the set of all such points is a circle. Moreover, it seems that the circle intersects the line $O_1 + tO_1O_2$ and can be computed as follows:
Let $$C=\frac{c_1O_1 + c_2O_2}{c_1 + c_2}$$, $$D=\frac{c_2O_2 - c_1O_1}{c_2 - c_1}$$.
The circle has its origin at $O=\frac{1}{2}(C+D)$ and its radius is $\frac{1}{2}\lvert \lvert CD\rvert \rvert$.
I am asking how could one arrive at the conclusion that the set of points forms a circle and how would one go about coming up with the formulae above. I have troubles interpreting formulae for $C,D$. I actually could make sense of $C$ as of a weighted average, but $D$ stupefies me. Could you help me please?
You could do it algebraically. Say $O_1 = (x_1, y_1)$ and $O_2 = (x_2, y_2)$. Then the points you're after are the points $(x, y)$ that satisfy $$ c_2\sqrt{(x-x_1)^2 + (y-y_1)^2} = c_2\sqrt{(x-x_2)^2 + (y-y_2)^2} $$ This is not too difficult to square and then rearrange to a standard circle equation. And it's even easier to see that it becomes a circle if you don't ned the explicit equation, since this becomes a quadratic equation in $x$ and $y$, and the coefficients of $x^2$ and $y^2$ are clearly the same, and there is no $xy$ term. Then, since we know there are at least two solutions (namely $C$ and $D$), the set of solutions to this quadratic is non-empty and must be a circle.
You could also do it geometrically, using angle bisectors. Take a point $X$ so that $|O_1X| = \frac{c_1}{c_2}|O_2X|$ with $c_1\neq c_2$. Then since $$ \frac{|O_1X|}{|O_2X|} = \frac{|O_1C|}{|CO_2|} $$ we get that $CX$ is an angle bisector of $\angle O_1XO_2$ by the angle bisector theorem. Similarily, we get that $XD$ is an exterior angle bisector of the triangle $\triangle O_1XO_2$ at $X$. Thus $\angle CXD =\frac{180^\circ}{2}=90^\circ$, so $X$ must lie on the circle with $CD$ as diameter.