Can anyone solve this geometric construction problem?

In following analysis, we will assume $R$ is lying outside the circle.

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Summary

• Even assume solutions exist. In general, $D$ is not (classically) constructible.
  (i.e by compass and straightedge alone)

• Algebraically, finding $D$ is equivalent to solving a quartic polynomial.

• Geometrically, one can use the intersection between a pair of hyperbola and circle to determine $D$. i.e. $D$ is conic constructible.

Part I - Finding $D$ is equivalent to solving a quartic polynomial

Choose a coordinate system such that:

• $R$ is the origin $O = (0,0)$,
• the line $\ell$ at hand is $x = 1$,
• the circle $\mathcal{C}$ at hand is centered at $A = (x_A,y_A) = (c\cos\theta,c\sin\theta)$ with radius $b$.

We will assume $c > b$, i.e. the circle doesn't contain the origin. Let

• $a^2 = c^2 - b^2$.
• $D = (1,t)$ be a point on line $\ell$,
• $E,F$ be the intersection between the line $OD$ and circle $\mathcal{C}$.
• $M$ be the midpoint of $EF$.

The line $OD$ will be described by the equation $y - xt = 0$ and its distance to $A$ equals to

$$|AM| = \frac{|y_A - x_A t|}{\sqrt{1+t^2}}$$

If $t$ is chosen such that $|OD| = |EF|$, we will have $$ |AE|^2 = |AM|^2 + |ME|^2 = |AM|^2 + \frac14 |EF|^2 = |AM|^2 + \frac14 |OD|^2 $$ This leads to $$ \frac{(y_A - x_A t)^2}{1+t^2} + \frac14 (1+t^2) = b^2 \;\iff\; (1+t^2)^2 - 4b^2(1+t^2) + 4(y_A - x_At)^2 = 0 \tag{*1} $$ So finding $D$ is equivalent to solving a quartic polynomial.

Part II - In general, $D$ is not classically constructible.

Consider the case $A = (2,1)$ and $b = 1$. The equation on RHS of $(*1)$ becomes $$t^4+14t^2-16t+1 = (t-1)(t^3+t^2+15t-1) = 0$$

There are two solutions, $t = 1$ and $t = t_0$ where $t_0$ is a root of $p(t) = t^3 + t^2+ 15t−1$.

Since all coordinates, distances and radii are integers, if there is a general construction of $D$ by compass and straightedge alone, then $t_0$ will be a classically constructible number. There will be a field extension $K$ of $\mathbb{Q}$ with $t_0 \in K$ and $[ K : \mathbb{Q} ]$, the dimension of $K$ over $\mathbb{Q}$, equals to $2^n$ for some integer $n$. Since $p(t)$ is irreducible, this leads to

$$2^n = [ K : \mathbb{Q} ]= [ K(t_0) : \mathbb{Q} ] = [ K(t_0) : \mathbb{Q}(t_0) ][\mathbb{Q}(t_0) : \mathbb{Q} ] = 3 [ K(t_0) : \mathbb{Q}(t_0) ]$$ which is impossible.

This rules out the possibility for a general construction of $D$ by compass and straightedge alone.

Part III - Locate $D$ with help of a pair of hyperbola and circle.

Consider the point $\displaystyle\;B = (u,v) = \left( \frac{2ab}{1+t^2}, \frac{2abt}{1+t^2} \right)$ which lies on the line $OD$. Since

$$u^2 + v^2 = \frac{4a^2b^2}{1+t^2} = 2ab u \quad\iff\quad (u-ab)^2 + v^2 = (ab)^2$$

$B$ is lying on a circle centered at $X = (ab,0)$ passing through $O$.

Construct another coordinate system by rotating the axis for an angle $\theta$. In the new coordinate system, $A$ is lying on the new $x$-axis and the coordinates of $B$ is give by $$(\tilde{u}, \tilde{v}) = (u\cos\theta + v\sin\theta, -u\sin\theta + v\cos\theta)$$ In terms of $(\tilde{u},\tilde{v})$, the equation on LHS of $(*1)$ becomes

$$\frac{(y_A - x_A t)^2}{(1+t^2)^2} + \frac14 = \frac{b^2}{1+t^2} \iff \frac{(y_A u - x_A v)^2}{4a^2b^2} + \frac14 = \frac{u^2+v^2}{4a^2}\\ \iff \frac{\tilde{u}^2+\tilde{v}^2}{a^2} - \frac{c^2\tilde{v}^2}{a^2b^2} = 1 \iff \frac{\tilde{u}^2}{a^2} - \frac{\tilde{v}^2}{b^2} = 1 $$

This is the equation of a hyperbola with foci $(\pm c,0)$ and vertices $(\pm a, 0)$ in $(\tilde{u},\tilde{v})$-plane.
Convert it back to $(u,v)$ coordinate, this define a hyperbola with foci at $\pm A$, center at $O$ and semi-major axis $a$.

This means we can locate $D$ by first finding $B$ as the intersection of a hyperbola and a circle and then intersect $OD$ with line $\ell$.

Part IV - $D$ is conic constructible.

As a result of this, there exists a general procedure to construct $D$ using conics.

The picture below is an instance for such a construction. The line $\ell$ is colored in red and the circle $\mathcal{C}$ is colored in orange. The procedure is given after the picture.

1. Construct a line through $O$ tangent to circle $\mathcal{C}$. Let $T$ be the point of tangency.
2. Construct a circle through $T$ centered at $O$, let the circle intersect line $OA$ at $S$.
3. Construct $A'$, the mirror image of $A$ with respect to $O$.
4. Construct hyperbola $\mathcal{H}$ (the one colored in cyan) through $S$ with with foci at $A$ and $A'$.
5. Construct a line $\ell'$ through $O$ perpendicular to $\ell$.
6. Start from $T$, construct a bunch of parallel lines, points and circle and locate the point $X$ on $\ell'$ with $|OX| = ab$.
7. Construct a circle $\mathcal{C}'$ (the one colored in magenta) through $O$ centered at $X$.
8. Let $B_1$ and $B_2$ be the two intersections of hyperbola $\mathcal{H}$ and circle $\mathcal{C}'$.
9. Let $D_1$ and $D_2$ be the intersection of $\ell$ with the two lines $OB_1$ and $OB_2$.
   They are the points you are looking for.

Okay. Case 1. The point and center of circle are on opposite sides of the Line.

Orient a coordinate system so that the point R is (-b, 0), the line is x=0, the circle has radius 1 and is centered at (w,z).

Construct a line passing through point R with an angle to the x-axis of $\alpha$. This line will intersect x= 0 at D =(0, b $\tan \alpha$). DF = $b\sec \alpha$. The equation of the line is $y = (x+b)\tan \alpha$.

The equation for the circle is $(x - w)^2 + (y- z)^2 = 1$.

So solve for $(x - w)^2 + ((x + b)\tan \alpha - z)^2 = 1$ for the two points of intersection. Call $E = (x_E, (x_E + b)\tan \alpha)$ and $F = (x_F, (x_F + b)\tan \alpha)$ where $x_E$ and $x_F$ are the two solutions.

Then solve for $\alpha$ where RD = $b\sec \alpha = EF = \sqrt{(x_E - x_F)^2 + (x_E - x_F)^2 \tan^2 \alpha}$

Case 2: similarly.