Three circles have the same radical axis? 
Given three circles $\bigcirc O_1$, $\bigcirc O_2$, $\bigcirc O_3$, let $A$, $B$, $C$ be three points on $\bigcirc O_3$. If we have 
  $$
\frac{\operatorname{power}(A, \bigcirc O_1)}{\operatorname{power}(A, \bigcirc O_2)}=
\frac{\operatorname{power}(B, \bigcirc O_1)}{\operatorname{power}(B,\bigcirc O_2)}=
\frac{\operatorname{power}(C, \bigcirc O_1)}{\operatorname{power}(C, \bigcirc O_2 )}$$ (where $\operatorname{power}(P, \bigcirc Q)$ denotes the power of point $P$ with respect to $\bigcirc Q$), can we conclude that these circles have the same radical axis?

 A: I'll change your notations to make it more comfortable for me.
Let $\mathscr C_i$ be the circle centered at $O_i$ with radius $r_i$, for $1\le i\le 3$.
Denote by $P(X,\mathscr C)$ the power of point $X$ to circle $\mathscr C$.
Let $\alpha$ such that $\alpha=\frac{P(X,\mathscr C_1)}{P(X,\mathscr C_2)}$ for $X\in\{A,B,C\}$.

Claim: If $A,B,C$ are three distinct points, then for any $X\in\mathbb R^2$, we have
$P(X,\mathscr C_1)=\alpha P(X,\mathscr C_2) + (1-\alpha) P(X,\mathscr C_3)$.
In particular, if $\alpha\notin\{0;1\}$ we easily deduce that the three circles have the same radical line. (When $\alpha\in\{0;1\}$ we only have two distinct circles so the result actually still holds.)

Proof: To exploit the fact that $A,B,C$ are on $\mathscr C_3$, we re-express the power of a point:
\begin{align*}
P(X,\mathscr C_i) ~=~ \|X-O_i\|^2-r_i^2
~=~ \| X-O_3\|^2-2\langle X-O_3,\ O_i-O_3\rangle+P(O_3,\mathscr C_i)
\end{align*}
When $X\in\{A,B,C\}$ we additionally have $\|X-O_3\|^2=r_3^2$ and
$P(X,\mathscr C_1)=\alpha P(X,\mathscr C_2)$. Using the expression above we get
$$(1-\alpha)r_3^2+P(O_3,\mathscr C_1)-\alpha P(O_3,\mathscr C_2) ~=~ 2\big\langle X-O_3,\ (O_1-O_3)-\alpha (O_2-O_3)\big\rangle$$
The left hand side is independant from $X$, this implies that the right-hand side must be constant regardless of what $X\in\{A,B,C\}$ we pick.
Because $\|X-O_3\|=r_3>0$ and assuming that $A,B,C$ are distinct, this implies $$O_1-O_3 ~=~ \alpha (O_2-O_3)$$
and
$$P(O_3,\mathscr C_1) ~=~ \alpha P(O_3,\mathscr C_2) -(1-\alpha) r_3^2$$
We can then re-inject those two identities into $P(Y,\mathscr C_1)$ for an arbitrary $Y\in\mathbb R^2$:
\begin{align*}
P(Y,\mathscr C_1) &= \|Y-O_3\|^2 -2\langle Y-O_3,\ O_1-O_3\rangle +P(O_3,\mathscr C_1) \\
&= \| Y-O_3 \|^2 -2\alpha\langle Y-O_3,\ O_2-O_3\rangle +\alpha P(O_3,\mathscr C_2) -(1-\alpha)r_3^2 \\
&= \alpha P(Y,\mathscr C_2) +(1-\alpha) P(Y,\mathscr C_3)
\end{align*}
