Construct point on a circle such that the reflection in that point is horiztonal

Let $P$ be a point in the plane outside the unit circle. There is a unique point $Q$ on the circle such that a light ray from $P$ is reflected in the circle at $Q$ and emerges parallel to the $x$-axis. Is it possible to construct $Q$ using a ruler and compass?

If $P$ is a point on the piece of paper in this picture, the image of $P$ will appear to an observer at infinity to be at a point easily constructed from $Q$.

I would also be interested in a proof that such a construction is impossible.

Solution 1:

Starting from @Narasimham's nice solution, his/her last equation can be solved analytically.

Using the tangent half-angle substitution $\theta_Q=2 \tan ^{-1}(t)$, the equation write $$y_P \,t^4+ 2( a+2 x_P)\,t^3-6y_P\, t^2+2 ( a-2 x_P)\,t+y_P=0$$ which can be solved with radicals.

"More than likely", this equation has two distinct real roots and two complex conjugate non-real roots since $$\Delta=-256\left(a^6+3\left(5 y^2-4 x^2\right)a^4+48 \left(x^2+y^2\right)^2a^2-64 \left(x^2+y^2\right)^3 \right)$$

Solution 2:

Yes, we can. It seems to me a locus finding problem in reflection geometry. Incident ray direction and incident ray segment length can be taken $2 \theta$ anti-clockwise and constant $b$ respectively when reflected ray QR is horizontal.

Directly from the figure at left parametric equation of locus of P of two components from $(a=1,b)$ is the spiral

$$(x_P,y_P)=(a\cos\theta_Q+b\cos 2\theta,\,a\sin \theta_Q + b\sin 2\theta_Q)\tag 1$$

which ( appearance similar to Limacon ) is an algebraic curve of fourth degree (not converted here to cartesian coordinates ) sketched below:

When $b$ segment length is perturbed the yellow band is obtained with rulings of all possible incident rays.

For Ruler and Compass choose a point R, draw a line parallel to x-axis to cut mirror M at Q which is the unique point. Drawing a perpendicular to OQN, transferring angle $\theta$ to left of normal OQN by cutting off arc length QR with compass to find P, is simple construction.

EDIT1:

If the above is an indirect method we can directly find coordinates of Q:

Using above parametrization and from the condition that the angular bisector of QP, QO is QR we arrive at the required unique $\theta_Q $ implicit condition to be satisfied for given P$(x_P,y_P).$ Calculation detail is omitted.

$$y_P \cos 2 \theta_Q- x_P \sin 2 \theta_Q + a \sin \theta_Q =0 \tag2 $$

Numerical solution by Newton-Raphson can determine $\theta_Q$.