Given two circles $S_1$ and $S_2$, a line $l_1$, and a length $a$ that is less than the sum of the diameters of the circles, construct a line $l$, parallel to $l_1$, so that the sum of the chords that $S_1$ and $S_2$ cut from $l$ is $a$. (You may assume that the interiors of circles $S_1$ and $S_2$ are disjoint, and that the circles are positioned so that at least one such line $l$ exists.)

I tried working backwards, and translating the circles such to connect the two chords, but I did not get anything useful from that. Can I have a little hint as to how to start/procede from my current position? Thanks!


let me see if i can construct the required line geometrically.

(a) draw lines $s_1, s_2$ orthogonal to line $l_1$ through the centers $O_1, O_2$ of $S_1$ and $S_2$ respectively. let $d$ be the distance between the parallel lines constructed.

(b) construct the length $d- {1\over 2} a$ (added later: this is the gap between the two circles in the direction of $l_1$ needed.)

(c) construct a point $O$ so that $O$ is between lines $s_1, s_2$ and $O_1O = d - {1 \over 2}a$

(d) draw a circle $S$ centered at $O$ and with the same radius as $S_1.$

(e) find the points where $S$ and $S_2$ intersect.

(f) draw lines through the points found in (e) parallel to line $l_1$.

the line/s in (f) is the answer to your problem.


Here is a graphic to better understand the construction. Point $A$ and the dotted circle were not mentioned in abel's description: they are used to find the point $O$.

enter image description here

| cite | improve this answer | |
  • $\begingroup$ I tried this in Geogebra, and as long as the input sizes are within certain limits, this works! (1) Could you show the justification for this? (2) May I add a graphic of my verification to clarify your construction? +1 and this answer should be accepted! $\endgroup$ – Rory Daulton Dec 23 '14 at 14:27
  • $\begingroup$ @RoryDaulton, thanks for your verification. please add the figures from geogebra. i don't have and don't know how to add figures. adding figures would certainly make it easier to follow my reasoning. anyway, i will add some justification. $\endgroup$ – abel Dec 23 '14 at 14:42

NOTE: This post has been edited due to @RicardoCruz catching a mistake. I also changed the desired variable from $z$ to $u$, for my own reasons.

Here is an approach that leads to a not-so-easy but possible construction.

Rotate your circles and line onto a Cartesian frame of reference so line $l$ is horizontal and translate so circle $S_1$ is centered at the origin with radius $s$. Its equation is then $x^2+y^2=s^2$. Let's then say that circle $S_2$ is centered at point $(c,d)$ and has radius $r$. (Reflections could be made to make both $c$ and $d$ nonnegative.) Let's finally say that desired line $l_1$ has the equation $y=u$. Given $a$, $c$, $d$, $r$, and $s$, we want to find $u$ so the sum of the chords of line $l$ with circles $S_1$ and $S_2$ is $a$.

Two circles with chords

We can easily find that the length of the chord in $S_1$ is $2\sqrt{s^2-u^2}$ and the length of the chord in $S_2$ is $2\sqrt{r^2-(d-u)^2}$. Therefore, the equation we want to solve for $u$ is


Removing the second square root by getting it alone and squaring leads to

$$\left(\frac{a^2}4+d^2+s^2-r^2 \right)-2du=a\sqrt{s^2-u^2}$$

I am too lazy to continue, but squaring this equation gives a quadratic equation in $u$. The solution (or solutions: there may be two valid ones!) can be constructed with compass and straightedge, which can easily be used to construct line $l_1$. Constructing the solution(s) for $u$ looks messy but there may be some tricks to make it easier.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your input. I'm just wondering, since I can't prove it, is there any way to find the distance between the two chords in terms of the lengths of the two chords? $\endgroup$ – Pakquebchsoflwty Dec 20 '14 at 18:27
  • 1
    $\begingroup$ There is something that doesn't fit in the equation you got. I think the second term of the RHS should be $2 \sqrt{r^2-(d-z)^2}$. If you correct it, you will get a quadratic equation. $\endgroup$ – RicardoCruz Dec 20 '14 at 19:36
  • $\begingroup$ @RicardoCruz: You are absolutely right! (And I was so wrong!) Thanks for the catch. I'll edit my answer immediately. How can I give you some of my (as-yet-non-existent) reputation points for this answer? $\endgroup$ – Rory Daulton Dec 20 '14 at 21:20
  • $\begingroup$ @Pakquebchsoflwty: Note that I made major changes to my answer. You problem is indeed solvable with compass and straightedge. $\endgroup$ – Rory Daulton Dec 20 '14 at 21:36
  • $\begingroup$ @RoryDaulton You're welcome. I'm happy to help. $\endgroup$ – RicardoCruz Dec 20 '14 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.