This is a variation of Alhazen's Billiard Problem.
Suppose we have a semicircular billiards table of radius r centered at the origin O, and a billiard ball placed somewhere on the 'x-axis' of the table. Let us call this point P, with coordinate $(0,p)$, with the stipulation that $0$ < P < r. Let the distance OP be $p$. On what point on the table should we aim at such that the billiard ball will bounce off the edge of the table once and into the other 'end' of the x-axis at $(0,-r)$?
Is it also possible to generalize this for any point $(x,y)$ in the circle?
Thanks.
Let $P=(p,0)$ be the starting point, $E=(-r,0)$ the end point, $Q$ the bouncing point and $\theta=\angle POQ$. Then we have $PQ^2=p^2+r^2-2pr\cos\theta$, $EQ^2=2r^2(1+\cos\theta)$ by the cosine rule, and $PO:OE=PQ:EQ$, because $QO$ is the bisector of $\angle PQE$. Combining these we get $$ p^2:r^2=(p^2+r^2-2pr\cos\theta):(2r^2(1+\cos\theta)) $$ whence $$ \cos\theta={r-p\over2p}. $$ Notice that a solution exists only if $p\ge r/3$.