Billiards in a circular table

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.

• When you say semicircular billards table, do you really mean a circular billards table? Or at least, is the center of what would be the circle at the origin, or is the center of mass of the semicircle (a different 'center') at the origin? – NoseKnowsAll Sep 17 '15 at 15:56
• Sorry for the confusion; yes, I meant the centre of what be the circle at the origin. The reason for a semicircular table as opposed to a circular table is because I am looking for solutions to a closed path with 3 bounces. If the billiard bounces back into the x-axis at the edge of the circle, by symmetry the ball should return to its starting position, and thus consideration of the other half of the circle is not needed. – Razorlance Sep 17 '15 at 16:22

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 $p\ge r/3$.
• From your comment above I supposed point $P$ was in fact a generic point inside a circle, that you divided by convenience into two halves. So I don't understand your question: are you asking for a generic position of $P$ inside a fixed semicircle, with the ball bouncing once and hitting one of diameter ends? – Aretino Sep 19 '15 at 14:48
• Ah, sorry for not explaining further. Yeah, I'm asking for a generic position of $P$ in a fixed circle, such that the diameter is no longer the x-axis but some arbitrary diameter of the circle. – Razorlance Sep 19 '15 at 14:52
• $x_Q$, the x-coordinate of Q, is equal to $r\cos\theta$, right? – Razorlance Nov 28 '15 at 8:40
• And what if $p$ was negative? – Razorlance Nov 28 '15 at 8:56