Let $\omega_1$ and $\omega_2$ be circles with respective centres $O_1$ and $O_2$ and respective radii $r_1$ and $r_2$. Let $k\in\mathbb{R}\setminus\{1\}$. Prove that the set of points $P$ such that $$PO_1^2-r_1^2=k(PO_2^2-r_2^2)$$ is a circle.
It is clear to me how to do this algebraically. If I let $P$ be $(x,y)$, assume without loss of generality that $O_1$ is $(0,0)$ and let $O_2=(a,b)$, then $$x^2+y^2-r_1^2=k((x-a)^2+(y-b)^2-r_2^2),$$ so $$(k-1)x^2+(k-1)y^2-2kax-2kby+ka^2+kb^2-kr_2^2+r_1^2=0$$ and so on.
But is there an easy way of proving this geometrically?
Note that the given condition is equivalent to $\dfrac {\sqrt {PO_1^2-r_1^2}}{\sqrt {(PO_2^2-r_2^2)}} = \sqrt {k}$, a constant.
This satisfies the condition of a circle, the Apollonius circle.
Picture added:-