Assume that $l$ is a line and $A,B,C,D$ are points in general position. How many pairs of circles $w_1,w_2$ are there such that $w_1$ goes through $A,B$, $w_2$ goes through $C,D$ and the radical axis of $w_1,w_2$ is $l$?
By writing the equations, I think the answer is there is at most one such pair. can anyone give a geometric proof of this fact?
Suppose that we have circles $\omega_1$ and $\omega_2$ with required properties. Let $AB$ and $CD$ cut $\ell$ at $E$ and $F$, respectively. Since $\ell$, $A$, $B$, $C$, $D$ are in general position, $E$ and $F$ exist and $E\neq F$.
Let $G$ be the second intersection of $AF$ and $\omega_1$ (if $AF$ is tangent to $\omega_1$ then take $G=A$). Power of the point gives $$\overrightarrow{FA} \circ \overrightarrow{FG} = \overrightarrow{FC} \circ \overrightarrow{FD}.$$ Note that this equality characterizes $G$ uniquely, i.e. this is the only point $X$ on the line $AF$ satisfying $\overrightarrow{FA} \circ \overrightarrow{FX} = \overrightarrow{FC} \circ \overrightarrow{FD}$. Since $E\neq F$, we have $G \neq B$. If $G\neq A$ then $\omega_1$ is the circumcircle of the triangle $ABG$. If $G=A$ then $\omega_1$ is the circle passing through $B$ tangent to $AF$ at $A$. In both cases, the circle $\omega_1$ is unique. Similarly we can show that $\omega_2$ is unique.