\PGset[0.8em]
\begin{picture}(16.5,11)

\drawline(1,4.5)(4.643,4.5)(1.893,1.15) % M, m=1.218

\drawline(6,4.5)(12,4.5) % AB
\dashline[80]{0.2}(9,0.846)(15,8.154) % EB

% F = 9,4.5
% O = 9, 6.963
\dashline[80]{0.2}(9,4.5)(9,6.963)(12,4.5) % FOB, OB = 3.882

% Ellipse:  u = 9.0  v = 6.963  a = 3.882  b = 3.882  phi = 0.0 Grad
\qbezier[20](12.882, 6.963)(12.882, 8.571)(11.745, 9.708)
\qbezier[20](11.745, 9.708)(10.608, 10.845)(9.0, 10.845)
\qbezier[20](9.0, 10.845)(7.392, 10.845)(6.255, 9.708)
\qbezier[20](6.255, 9.708)(5.118, 8.571)(5.118, 6.963)
\qbezier[20](5.118, 6.963)(5.118, 5.355)(6.255, 4.218)
\qbezier[20](6.255, 4.218)(7.392, 3.081)(9.0, 3.081)
\qbezier[20](9.0, 3.081)(10.608, 3.081)(11.745, 4.218)
\qbezier[20](11.745, 4.218)(12.882, 5.355)(12.882, 6.963)

% given AB and circle O, there is only one point K on O s.t. AKB = M = 50.6deg
% how do I find that point?

% 50.6 = atan((x-6)/(y-4.5)) + atan((12-x)/(y-4.5))
\dashline[80]{0.2}(6,4.5)(10.86,10.37)(12,4.5) % AKB

\put( 5.4, 3.8){$\scriptstyle A$}
\put(11.8, 3.8){$\scriptstyle B$}
\put( 9.9, 1.4){$\scriptstyle E$}
\put( 8.6, 3.8){$\scriptstyle F$}
\put( 8.8, 7.1){$\scriptstyle O$}
\put( 3.2, 3.8){$\scriptstyle M$}
\put(10.9,10.4){$\scriptstyle K$}


\end{picture}
\PGrestore