A friend of mine taught me the following question. He said he found it on a book a few years ago. Though I've tried to solve it, I'm facing difficulty.

Question: You know on a plane there is an invisible circle whose radius is less than or equal to $1$. Fortunately, you have already found that the lengths of the chords of a circle by two lines $l_1, l_2$ are $d_1, d_2$ ($2\gt d_1\ge d_2\gt 0$) respectively. By drawing another line, let's find this circle. If the line you'll draw crosses a circle at two points, then you'll get the length of the chord of a circle by the line. If the line you'll draw and a circle come in contact with each other, then you'll get the coordinates of the point of contact instead of getting $0$ as the length of the chord. If the line you'll draw neither crosses nor comes in contact with any circle, then you'll be able to draw another line just once more. Find the coordinates of the center of a circle.

This is all the question says. Could you show me how to find the coordinates?

This question has been asked previously on math.SE without receiving any complete answer: https://math.stackexchange.com/questions/468324/finding-an-invisible-circle-by-drawing-another-line

The $l_1\parallel l_2$ case has been already solved (see Blue's answer on math.SE). On the other hand, the $l_1\not \parallel l_2$ case has not been solved yet.

I'm going to write several things about the $l_1\not \parallel l_2$ case which we've already found on math.SE. For further details, please see the page on math.SE.

In the following, suppose that $l_1:y=x\tanθ$, $l_2:y=-x\tanθ$ for $0<θ<\pi/2$ and that $a=\frac{d_1}{2}, b=\frac{d_2}{2}$.

1. Taking $l_3:y=0$ ($l_4:x=0$ if needed), then we can get two possible coordinates as the center of a circle. However, it seems difficult to decide just one coordinates because each line is symmetric about the origin. Hence, a new line, which is not $y=0$, is needed as $l_3$.

2. We can represent every possible invisible circle as the following: $$C_{\pm+}:\left(x-\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2$$ $$C_{\pm-}:\left(x-\frac{-d-{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d-{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2$$ where $d$ satisfies the following: $$-\sqrt{1-a^2}\le d\le \sqrt{1-a^2}.$$

Hence, we know that the center of each circle is on the following hyperbola: $$xy=\frac{a^2-b^2}{4\cosθ\sinθ}.$$ if $d_1-d_2>0$.

3. Blue got a quartic in $h$ with $\phi, p, c$ supposing that $l_3:x\sin\phi−y\cos\phi+p=0$ cuts a chord of length $2c$ in the circle with center $(h,k)$ and radius $r$.

4. In Blue's quartic, taking $\phi=0, \frac{\pi}{2}$ don't work in general because such lines don't necessarily hit every circle in a given sub-family of circles.

Update: I'm going to write my idea. I hope this would be helpful.

First, let's call the following circles 'upper-right sub-family of circles': $$C_{\pm+}:\left(x-\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2.$$ Also, let's call the following circles 'lower-left sub-family of circles': $$C_{\pm-}:\left(x-\frac{-d-{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d-{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2$$ where $d$ satisfies $-\sqrt{1-a^2}\le d\le \sqrt{1-a^2}$.

We can see that each center of upper-right sub-family is in the first quadrant, and that each center of lower-left sub-family is in the third.

I've been looking for a special line $L$ which satisfies the following three conditions. If we can find such line, we can take the line as $l_3$.

1. $L$ crosses every circle of upper-right sub-family.

2. $L$ never crosses any circle of lower-left sub-family.

3. Each length of the chord of each circle of upper-right sub-family by $L$ is different from each other.

Now let $l_3$ be $x\sin\phi-y\cos\phi+p=0$. Actually, we can write these three as the following:

1. $|\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\sin\phi-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\cos\phi+p|\lt \sqrt{d^2+a^2}$ for any $d$.

2. $|\frac{-d-{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\sin\phi-\frac{d-{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\cos\phi+p|\gt \sqrt{d^2+a^2}$ for any $d$.

3. $f(d)=\sqrt{(d^2+a^2)-|\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\sin\phi-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\cos\phi+p|^2}$ is monotone increasing or decreasing function.

Then, we can say that if there exists a set of $(\phi,p)$ which satisfies these three conditions, then we can take $l_3$ to be the line $x\sin\phi-y\cos\phi+p=0$.

If this line doesn't cross any circle, then we can take $l_4$ to be the line which is origin-symmetric to $l_3$.

However, I neither know if there exists such $(\phi,p)$ for any $(\theta, a, b)$, nor know how to get such $(\phi, p)$ if it exists.

  • $\begingroup$ please forgive my question, but did I understand the question right, that an equation of the lines and the length of the chords on the respective line is given? If that is the case, I have an idea, which could check. Maybe I'm asking too much, but a sketch of the problem would be nice. $\endgroup$ Commented Aug 29, 2013 at 8:45
  • $\begingroup$ @ManfredWeis: I'm going to write 'my understanding'. Let's imagine that you are the person who is going to find the coordinates and that you are going to draw a line $l_1$(Note! not $l_3$). You draw a line called $l_1$ somewhere on a plane(You know nothing about an invisible circle, so all you can do is to draw the first line somewhere). Then, fortunately you get a value $d_1$. Then, you know $l_1$ crossed an invisible circle at two points(This is all you can know.) Next, when you draw another line called $l_2$, again fortunately you get a value $d_2$. (continued) $\endgroup$
    – mathlove
    Commented Aug 29, 2013 at 9:41
  • $\begingroup$ (part2) Again you know $l_2$ crossed a circle at two points. (note that before drawing $l_2$ you have two choices. One is to draw $l_2$ parallel to $l_1$. The other is to draw $l_2$ not parallel to $l_1$. Also, in the former, you can decide how far $l_2$ is from $l_1$. In the latter, you can decide where and what angle $l_2$ crosses $l_1$.) Then, we can supporse the following: In the former, we can suppose that $l_1:y=t, l_2:y=-t$ for some $t\gt 0$. In the latter, we can suppose that $l_1:y=x\tan\theta, l_2:y=-x\tan\theta$ for some $0\lt \theta \lt \pi/2$. Blue's answer would be helpful. $\endgroup$
    – mathlove
    Commented Aug 29, 2013 at 9:44
  • $\begingroup$ thanks for the quick reply; I guess, my idea could work. $\endgroup$ Commented Aug 29, 2013 at 10:34
  • $\begingroup$ @ManfredWeis: Could you share your idea with us? $\endgroup$
    – mathlove
    Commented Aug 30, 2013 at 14:17

3 Answers 3


This I think is the big clue: " If the line you'll draw and a circle come in contact with each other, then you'll get the coordinates of the point of contact instead of getting 0 as the length of the chord."

If the first line cuts a chord of length $d_1$ then the circle is enclosed in a band centred on the line, ranging from being centred on the line when the circle has diameter $d_1$ to a unit circle offset on either side of the line, and the lines bounding the band are tangents to both these unit circles.

With that in mind, what you should do is draw the third line parallel to the first at a distance of $\frac{d_1}{2}$ from it. If the circle is offset the other side of the line, you then have a second chance, and you draw another parallel line the same distance the other side.

One of these lines will then be either tangent to the circle (if it is centred on the line and has diameter $d_1$), and in this case the coordinates of the point of contact easily allow one to deduce the circle's centre.

Otherwise one of these lines will cut out a chord, and the length of this combined with the original chord length and the distance apart of the parallel lines will allow the radius of the circle to be determined.

But once you know the circle's radius, the chord lengths cut by any two oblique lines allow its centre to be determined.

Very nice problem!

  • $\begingroup$ @John R Ramsden: Thank you for answering this question. I think the answer is not correct. As you wrote, we can get the radius. However, your last sentence is not correct because there are two possible circles which have the same radius and same chord lengths. See the picture Blue drew. How do you distinguish these two circles in spite of the fact that we only know lengths? I think your way leads two possible coordinates. $\endgroup$
    – mathlove
    Commented Aug 31, 2013 at 17:45
  • $\begingroup$ @mathlove: Flip, yes you're correct, and I don't see any way to rescue it! In that case the solution must be to draw your line through the intersection of the two starting lines at such an angle to ensure (given the chord lengths known initially) that the chord length of an intersection, if any, is unique to the side of the "cross" formed by the pair of initial lines, whatever the circle diameter. If there is no intersection, then try the same trick with the other pair of angles. (It's obvious one of these lines must clip the circle.) $\endgroup$ Commented Aug 31, 2013 at 19:02
  • $\begingroup$ It's clear a priori that $l_3$ must not go through the point $O$ where $l_1$ and $l_2$ meet, unless you can be sure that $l_3$ will be tangent to the hidden circle, because otherwise you cannot distinguish between the actual circle and its reflection about $O$. $\endgroup$ Commented Aug 31, 2013 at 19:17
  • $\begingroup$ @Noam: Drat, yes. It was rather presumptuous of me to assume a simple solution could be described, after the heroic efforts of people like Blue who have already replied (in the mathoverflow thread in his case). I'll say no more unless I can devise a formal solution! $\endgroup$ Commented Aug 31, 2013 at 20:15
  • $\begingroup$ P.S. squinting at Blue's figure, a line tangent to one of the extremal unit circles and intersecting the centre of the opposite unit circle might work - At least one knows where all these unit circles are, and that may be a chink in the thing's armour! $\endgroup$ Commented Aug 31, 2013 at 20:27

I'm not a specialist, but I'll give my idea. Call $O$ the intersection point between the two lines, $L$ and $L'$ the points on $l_1$ such that $OL<OL'$, $M$ and $M'$ the points on $l_2$ such that $OM<OM'$, (hence $d_1=LL'$ and $d_2=MM'$), $H$ the centroid of the system made of the points $L$, $L'$, $M$, $M'$, $C$ the center of the circle. Using Al-Kashi's theorem should allow to determine the coordinates of $C$.

  • 3
    $\begingroup$ I think the problem with this approach is that we do not know $L$, $L\prime$, $M$ or $M\prime$, only the lengths $d_1$ and $d_2$. And only knowing these lengths there will be a 1-parameter continuous family of circles satisfying the constraints. Even restricting to radius $\leq 1$, there are still (generically) infinitely many solutions. So the third line is necessary to find the final parameter locating the actual circle from within the continuous family. $\endgroup$
    – ARupinski
    Commented Aug 30, 2013 at 22:02

4please excuse my bad formatting and typesetting, but I'm currently under time pressure;
please feel free to downvote.

The following is a rather narrative description of a possible way to find a hidden circle under the stated condition that no parallel lines may be used in detecting it.
I do not provide formulas for the solution, but only a way how to proceed; the formulas could be obtained with the help of CAS and I may supply them in later edits; my apologies also for that initial shortcomming.

First lets discuss the task of choosing an $l_3$ that doesn't miss the circle;

simply guessing $l_3$ doesn't guarantee that it will ever hit the circle; this is so because there is no lower limit on the circles radius.

The next idea would be to try the first bisector of $l_1$ and $l_2$ and, if that fails, try the other bisector, that is orthogonal to the first one.
This strategy works in all cases, no matter how small the radius is.

But, we can do better by choosing $l_3$ in a way that guarantees that it hits the circle at first try.
To see this, one has to anticipate the implications of the condition that the radius is not greater than 1.
In fact, it is the mere existence of an upper bound and not the value $1$ itself, that is important here:

we take one of the unit circles, whose center is $sqrt(1-a^2)$ away from $l_1$ and $sqrt(1-b^2)$ from $l_2$ and, whose center is farthest from the intersection point $S$ of $l_1$ and $l_2$; if $S$ is inside that unit circle, then any choice of $l_3$ will do and we can chose it as the bisector of the larger angle formed by $l_1$ and $l_2$ (for better numerics in subsequent steps).
in the general case there will be two of such unit circles, that are point-reflections of each other w.r.t. $S$.
$l_3$ is then chosen as the line through $S$, that is tangent to that unit circle in a point on the shorter arc connecting the intersection points of $l_2$ with that unit circle.

Had there been no upper limit on the circle's radius, then the way in which $l_3$ has been constructed, would not have been possible, because the tangent could come arbitrarily close to $l_2$.

Having constructed $l_3$ as described, either yields a tangent or it intersects the circle in two points; lets assume the worse, it will become clear that both cases can be handled in the same manner.

To summarize: what we have so far are three distinct lines with a common intersection point and each of the lines intersects the unknown circle in at least 1 point. In the next step it will be explained, how to obtain candidate points for the circle's center, and finally, how to disambiguate the centers.

Suppose for a moment, that the radius of the circle is known; what could we accomplish with that additional information?
As we also have the length $d_1$ and $d_2$ and $d_3$ of the segments that the circle cuts out of $l_1$, $l_2$ and, $l_3$, we could then determine the circle's distance to the respective lines but, we couldn't determine on which side of a line the center lies. Therefore we have to intersect 3 pairs of parallel lines, that resemble all possible locations of the circle's center with respect to $l_1$, $l_2$ and, $l_3$.
The candidate-points for the circle's center are then those, in which 3 of the parallel lines intersect.
In the case that $l_3$ is tangent to the sought circle, the corresponding pair of parallel lines is identical to $l_3$, which has no impact on the calculation of the aforementioned intersections.

If the radius of the circle is unknown, then the unknown radius can serve for parameterizing the location of the center via 3 mutually non-parallel lines from the three pairs of parallels, which amounts to solving 3 equations, that are nonlinear in the distance of the center to $l_1$, $l_2$ or, $l_3$
(the 3 equations come from selecting different sides of a line for the center; actually there are 6 possible ways of choosing the center's side relative to the three lines, but there are 3 pairs of mutally equivalent side-combinations due to the symmetry w.r.t. the intersection of l1,l2 and, l3).

Finally the set of solutions has to be disambiguated by cleverly choosing additional lines; in the worst case, there could be four candidate centers and one must keep in mind that the resulting candidate circle's obtained from the previously calculated radii and centers, could be almost identical when all centers are very close to the intersection point $S$ and if the radii are fairly large.

The optimal strategy for disambiguation seems to be the following: use a line that is simultaneously tangent to two of the circles and a additionally intersects or touches a third one; if that is possible, the disambiguation doesn't need a further check and four lines suffice to find the circle's center.

Otherwise it can happen that the line didn't touch the sought circle, allowing us to exclude two of the candidates and requiring a further line to disambiguate the remaining two circles. Had the line instead touched one of the circles, it would have been the sought one and 4 lines would also have sufficed.

So in summary it can be said, that four lines suffice in the majority of cases but there are situations, where five lines are inevitable.


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