Locus of all midpoints. 
Two circles intersect at $A$ and $B$, $PAQ$ is a straight line through $A$ meeting the circles at $P$ and $Q$. Find a locus of a midpoint $PQ$.

Please give only hints and not the solution. This is a question of pure geometry and not analytic geometry.
 A: It is much harder to come up with a hint than answering the question.
Just stare at following picture before reading anything after it.
Let's see whether you can figure out the answer. 


Let $C$ and $D$ be centers of the circles and $O = \verb/mid/(C,D)$ be their midpoint.
Construct three lines perpendicular to line $PQ$, passing through $C, O, D$ respectively.
Let these three lines intersect line $PQ$ at $C', O'$ and $D'$.  
Since $A$ and $P$ lies on the circle centered at $C$. $|AC| = |PC|$ and $CC'$ is the perpendicular bisector for segment $AP$. This means $C' = \verb/mid/(P,A)$. By a similar argument, $D' = \verb/mid/(Q,A)$.  
Notice line $OO'$ is parallel to line $CC'$ and line $DD'$. Since $O = \verb/mid/(C,D)$,  we have 
$$O' = \verb/mid/(C',D') = \verb/mid/(\verb/mid/(P,A),\verb/mid/(Q,A))
= \verb/mid/(\verb/mid/(P,Q),A) = \verb/mid/(M,A)$$ 
Using the fact $OO'$ is perpendicular to $AM$, we get $|OA| = |OM|$.
From this, we can deduce the locus we seek is a circle centered at $O$ passing through $A$.
