Given two circles $C_1$ and $C_2$ with centre $O_1$ and $O_2$ respectively. Choose two arbitrary points, one on $C_1$ and one on $C_2$. We name them $P_1$ and $P_2$. Draw the line $P_1P_2$ and mark the midpoint, $M$ of the line. Draw two lines $O_1M$ and $O_2M$. Then, drop a perpendicular line from $P_1$ to $O_1M$ and a perpendicular line from $P_2$ to $O_2M$. How can I show that the intersection of the two perpendicular lines lies on the radical axis of two circles?
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1 Answer
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Let $S$ be te intersection of the perpendicular to $O_2M$ from $P_1$ and analogously define $T$. If one is able to prove that $STP_1P_2$ is a cyclic quadrilateral than it will be done. (Check yourself why)
This is easy because $MP_1=MS$ since $O_1M$ is the perpendicular bisector of $P_1 S$. Analogously $MP_2=MT$ and since $MP_1=MP_2$ by construction it follows that $STP_1P_2$ are equidistant from $M$ therefore the lie on the same circumference with centre $M$.
