Can anyone solve this geometric construction problem? I remember when I was in high school, one of my all-time favorite books was College Geometry by Nathan Altshiller-Court. Some of its problems kept me wondering for days and even weeks. Now after about 16 years, it is really strange for me that there was this one problem that I could never figure out, and it is still bugging me sometimes.
At the end of Chapter 1, the 25th review exercise says:

Through a given point $R$, draw a line cutting a given line in $D$ and a given circle in $E$, $F$ so that $RD=EF$.

Edit - Some notes:
Before any comments or answers, please note that this problem is tagged as Geometric Construction. So you must be familiar with these types of problems before giving this a try. For example, read this article and also take a look at this post.
Geometric Construction problems are more like mathematical puzzles and they often need little mathematical knowledge and high creativity.
 A: 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.
A: In following analysis, we will assume $R$ is lying outside the circle.

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.



*

*Construct a line through $O$ tangent to circle $\mathcal{C}$. Let $T$ be the point of tangency.

*Construct a circle through $T$ centered at $O$, let the circle intersect line $OA$ at $S$.

*Construct $A'$, the mirror image of $A$ with respect to $O$.

*Construct hyperbola $\mathcal{H}$ (the one colored in cyan) through $S$ with with foci at $A$ and $A'$.  

*Construct a line $\ell'$ through $O$ perpendicular to $\ell$.

*Start from $T$, construct a bunch of parallel lines, points and circle and locate the point $X$ on $\ell'$ with $|OX| = ab$.

*Construct a circle $\mathcal{C}'$ (the one colored in magenta) through $O$ centered at $X$. 

*Let $B_1$ and $B_2$ be the two intersections of hyperbola $\mathcal{H}$ and circle $\mathcal{C}'$.

*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.

A: The methods used so far ("classic constructions") are akin to trying to solve an equation by an algebraic method.
Many equations can not be solved algebraically, and for those we have developed a set of numerical methods: secant methods, interval bisection, Newton-Raphson etc.
This problem can also be approached as an iteration of constructions.
Step 1
Construct a tangent to the circle that passes through the point $R$. Call the point of contact with the circle $P_0$. Let $F_0=E_0=P_0$. Call the point of intersection of this line with the given line $D_0$.
Call the centre of the circle $P_1$. Construct the line through the point R and $P_1$. Call the points of intersection of this line with the circle $E_1$ and $F_1$. Call the point of intersection of this line with the given line $D_1$.
You now have two extremes:
${E_0F_0 \over RD_0}=0$
${E_1F_1 \over RD_1}={\text {diameter of circle} \over RD_1}$, which is going to be close to the maximum value possible. Yes, there are certain configurations of the line where this is not true, but it's a good rule of thumb.
If ${E_1F_1 \over RD_1}<1$, then we are unlikely to find another point such that ${EF \over RD}=1$ which is required if we are to have $EF=RD$.
If ${E_1F_1 \over RD_1}>1$, then we need to adjust our position slightly.
Step 2
Construct the line $P_0P_1$ (it's a radius) and find the point $P_2$ that divides $P_0P_1$ such that $P_0P_2:P_0P_1=1:{E_1F_1 \over RD_1}$.
Construct the line through $R$ and $P_2$. Call the points of intersection of this line with the circle $E_2$ and $F_2$. Call the point of intersection of this line with the given line $D_2$.
Measure and calculate ${E_2F_2 \over RD_2}$
If ${E_2F_2 \over RD_2}>1$, then find $P_3$ that divides $P_0P_2$ such that $P_0P_3:P_0P_2=1:{E_2F_2 \over RD_2}$.
If ${E_2F_2 \over RD_2}<1$, then find $P_3$ that divides $P_2P_1$ such that $P_2P_3:P_3P_1=1-{E_2F_2 \over RD_2}:{E_1F_1 \over RD_1}-1$.
Construct the line through $R$ and $P_3$. Call the points of intersection of this line with the circle $E_3$ and $F_3$. Call the point of intersection of this line with the given line $D_3$.
Measure and calculate ${E_3F_3 \over RD_3}$
Repeat.
This method has two advantages:
1) It's what you would do if you were faced with a drawing and had to find the point with a ruler. When I started looking at this problem I drew it in geogebra and moved my line around until the ratio of lengths was 1. 
2) It can be easily adapted to a spreadsheet. All the points of intersection can be readily calculated. Intersection of line with circle, line with line, even finding the tangent to the circle are all simple to do. 
Further adaptations:
I have used the ratio to identify the $P_2, P_3, ...$ but you could choose to go for an interval bisection instead - it might be slower... 
A: I attempted to find relation between constants  $ \alpha,\beta, c, h,R \
$ of two given lines and the circle described below. The idea was to device a geometric construction method that uses constants as given to find a method to construct but which in fact failed, but reporting the work nevertheless. 
Given lines and circle I  took in form wlog as: ( line D, straight line through origin and eccentric circle with E,F) as:
$$ y = \tan\alpha \cdot x - 2 c ;  \, y = \tan \beta\cdot x ; \, (x-h)^2 + y^2 = R^2  \tag{1} $$
For condition  the difference of roots of quadratic equation and radial ray segment projection lengths up to intersection should be equal. (Omitting worked out details, shall give it if  anyone is interested). The result is finally:
$$[ {{\dfrac {c \sec^2 \beta} {({\tan \alpha } - \tan \beta)} } } ]^2 + ( h\, \tan \beta ) ^2 = R^2 \tag{2}  $$
Solving for $ \tan \beta $ is tedious, Mathematica gives a humongous output. There are two real and two complex roots for the fourth order equation. Temporarily postponed for numerical solution but on the basis of relation (2) established upto this stage. 
Next found numerical solution and verified plots of the line segments formed as above are equal, showing that the inclinations of radial rays pf inclination $\beta$ is correct.
For the graph constants used are:
$$ (m = \tan \alpha ,c,h,R) = ( 0.8,3,8,5 ) $$

