A Guide to Finding the Perfect Vacation Spot

[slwarrior64]

 

[Opalg]

Rotate the diagram through $180^\circ$ about the midpoint of $MN$ to get two new circles, intersecting at say $A'$ and $B'$, so that $MANA'$ and $MBNB'$ are parallelograms. Use the alternate segment theorem in triangles $MAB$ and $NAB$ to show that angles $MBN$ and $MAN$ add up to $180^\circ$. Hence so do the angles $MBN$ and $MA'N$, and it follows that $MA'NB$ is a cyclic quadrilateral. So in the diagram below, the green circle (circumcircle of triangle $MAN$) and the red circle (circumcircle of triangle $MBN$) have the same radius: each of them goes to the other one when the diagram is rotated through $180^\circ$ about the midpoint of $MN$.
[TIKZ]%preamble \usetikzlibrary{calc,through}
\coordinate [label=above:$M$] (M) at (0,0) ;
\coordinate [label=above:$N$] (N) at (7,0) ;
\draw [thick] (-5,0) -- (10,0) ;
\node (P) [draw, thick, circle through=(M)] at (0,-5) {} ;
\node (Q) [draw,thick, circle through=(N)] at (7,-4.25) {} ;
\node (R) [draw, thin, circle through=(N)] at (7,5) {} ;
\node (S) [draw, thin, circle through=(M)] at (0,4.25) {} ;
\node [draw, red, circle through=(M)] at (3.5,-3) {} ;
\node [draw, green, circle through=(M)] at (3.5,3) {} ;
\coordinate [label=below: $A$] (A) at (intersection 2 of P and Q) ;
\coordinate [label=below: $B$] (B) at (intersection 1 of P and Q) ;
\coordinate [label=below: $A'$] (C) at (intersection 2 of R and S) ;
\coordinate [label=above: $B'$] (D) at (intersection 1 of R and S) ;
\draw [thick] (A) -- (M) -- (B) -- (N) -- cycle ;
\draw [thin] (C) -- (M) -- (D) -- (N) -- cycle ;[/TIKZ]

 

[slwarrior64]

Thank you so much!