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.
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.
Part I - Finding $D$ is equivalent to solving a quartic polynomial
Choose a coordinate system such that:
We will assume $c > b$, i.e. the circle doesn't contain the origin. Let
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.
They are the points you are looking for.