Here's the problem:

enter image description here

($2,5$ is $2.5$)

To determine $r$, I used Pythagoras and trigonometry to find that:

  • $\angle{BOC}=\dfrac{\beta}{r}$

  • $\tan{\dfrac{\beta}{r}}=-\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}$

As, from the graphic, $\angle{AOC}\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)$, when using $\arctan$, we get: $$\dfrac{\beta}{r}-\pi=-\arctan\left(\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}\right).\tag{1}$$

Plotting on WolframAlpha, an approximation of $r$ is $0.54$, which is what I get on Geogebra.

But I'm not satisfied. I relied on my eyes to know that the angle is $>\dfrac{\pi}{2}$, while if it belongs to $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, it's a different formula. It would be nice to be able to determine this for any given $\alpha$ and $\beta$ (where, of course, a circle like in the image would exist).

Do you know any other method? Like what I thought of is, if we manage to calculate one of the angles $\angle{ABC},\,\angle{AOC}$ or $\angle{ADC}$ (where $D\neq A$ is the other intersection of the line $(AB)$ and the circle), we'd be able to determine $r=\dfrac{\beta}{\angle{AOC}}$.

One other data that I got using tryhard analytic geometry: the coordonates of point $C$ are $\left(\dfrac{\alpha r}{r-\alpha},\dfrac{r\sqrt{(\alpha-r)^2-r^2}}{\alpha-r}\right)$.

Thank you in advance.


I just noticed that this always holds: $\angle{AOC}\in\left(\dfrac{\pi}{2},\pi\right)$. Thus $(1)$ always holds as long as $\beta$ is chosen in an adequate way.

So I guess here's the final result: given any construction as above, one has from $(1)$:


So, we look at $\beta$ as a continuous function of $r$ for now. Playing with Geogebra, I noticed that $\beta$ is increasing. If I'm not mistaken:

$$\beta'(r)=\pi-\arctan\left(\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}\right)+\dfrac{\alpha r}{\sqrt{(\alpha-r)^2-r^2}(\alpha-r)}$$

which is positive $(\arctan<\pi)$. This shows that $\beta$ is indeed increasing. Then we see that $$\lim_{r\to 0^+}\beta(r)=0$$ and $$\beta\left(\dfrac{\alpha}{2}\right)=\dfrac{\alpha\pi}{2}.$$

By the intermediate value theorem, this means that for any $\beta\in\left(0,\dfrac{\alpha\pi}{2}\right)$, we can make such a construction for exactly one possible $r$, which we can calculate numerically for specific examples.

  • $\begingroup$ If you have the coordinates of C, you can use the easily calculated midpoint of AB, say M, then MC is half of R. But I'm not sure how this would work with your expression for C involving R. Might be an approach to try though? Also possibly solve for angle COA by parametrizing the circle and using the known arc length for $\beta$ to get the upper bound of the path integral which is this angle? I think if you have angle COA you might be able to get the chord length CA and solve the isosceles triangle COA? Not sure these will work but maybe some ideas to try. $\endgroup$ – Circulwyrd Apr 23 '18 at 1:41
  • $\begingroup$ @Circulwyrd Thank very much for your suggestions. Please can you explain why MC is half of R? The idea of calculating the path integral is really nice! But I didn't manage to make an integral that involves $C$'s coordinates unless I get back to the equation above. But thank you! $\endgroup$ – Scientifica Apr 23 '18 at 8:58
  • $\begingroup$ @Scientfica My mistake. I meant if you have midpoints of BO and BC the distance between them is half of R (easily shown). But since you don't have O it wouldn't work! Another thing I noticed is if you compute a path integral based on independent var $t$ (so not using your expression for C), you just get the circle's curvature $\kappa$ for $\angle COA$ - as a result of the arc length being 1. But R still not eliminated. Another approach may use inversion - for example I believe if you construct circle center B radius BC it is orthogonal to the original circle via circle of inversion through O. $\endgroup$ – Circulwyrd Apr 24 '18 at 1:05
  • $\begingroup$ I see. Oh that's right. We know the coordinate formula (using r) for C's coordinates. So as you said we can find the coordinate formula of the point B' such as B is the inverse of B' through O. But I don't know how inversion may help. $\endgroup$ – Scientifica Apr 24 '18 at 10:23
  • $\begingroup$ It seems to me that it is $\angle COA$ that is $\frac{\beta}{r}$, so $\angle BOC$ is the complement? Could you give more detail for this step? $\endgroup$ – Circulwyrd May 8 '18 at 1:06

Given length $ \alpha= AB = 2.5 $. We can use two standard results: Tangents property (power) of a circle .. product of segments from external point $B$ is square of tangent length, and, the Pythagoras theorem before simplification:


(after sign error pointed out by OP)

$$ BC^2 = \alpha\cdot (\alpha - 2 r) = (\alpha-r)^2 -r^2 \rightarrow an \,identity \tag1 $$

By virtue of this identity and from what is given in the question no particular solution exists. I.e., the circle can then have any radius.

So we are free to choose radius $r$ entirely satifying the another basic arc length $ \beta=1$ (subtending arc $\angle AOC$ at center of circle):

$$ r+ r \sec(\pi-\beta/r) = r- r \sec(\beta/r) = 2.5 \tag2 $$

To improve computation accuracy the following equivalent equation ( removing inverse cos etc.) is adopted.

$$ \frac{1}{r}+ \sec^{-1} \frac{2.5-r}{r}=\pi \rightarrow \cos(\pi-1/r)(2.5-r)=r \tag3 $$

This transcendental equation can have only a numerical solution which is:

$$ r= 0.540472, \beta/r= 1/r= 106.011^{\circ} \approx 106^{\circ} \tag4, \angle CBO= 16 ^{\circ}$$

The geometric data is drawn to scale below:


  • $\begingroup$ Thank you for your answer. That's a nice tangent length formula, I didn't know it. The Pythagoras theorem gives $BC^2=(\alpha-r)^2-r^2$, not $r^2+(\alpha-r)^2$. $\endgroup$ – Scientifica May 6 '18 at 8:28
  • $\begingroup$ Thanks for your commment and apologize for the confusion due to error, so It is true for any or all cases of radius length. $\endgroup$ – Narasimham May 6 '18 at 15:47
  • $\begingroup$ No worries, made mistakes like those as well when tackling the problem. Could you please explain what your conclusion means in your last paragraph? $\endgroup$ – Scientifica May 6 '18 at 22:22
  • $\begingroup$ The constraint this does not take into account is the fixed arc length $\beta = 1$? I think this makes it such that the flexibility assumed that a tangent BC exists for any $r$ is not valid? $\endgroup$ – Circulwyrd May 7 '18 at 0:39
  • $\begingroup$ The secant formula looks good -- I did try plugging the values into geogebra to the same precision you give and unfortunately I get an angle of 107.1 for ∠AOC when I draw the tangent from (-2.5,0) to the given circle centered at (-0.569057,0) and touching the origin. Don't understand this -- see image at end of my answer showing the discrepancy. Can you explain it? Is it an issue with precision and/or is my drawing wrong? $\endgroup$ – Circulwyrd May 8 '18 at 3:18

I think this problem as stated may have no solution ! I'm not completely sure about this but it's easier to post as an answer so here goes.

In the diagram below the points Y(-2.5,0), O(0,0) are as in your image (you have B instead of Y, A instead of O). The ellipse $p$ is a locus of arc endpoints from O of length 1, with the curvature varying (as complete circles they would be described by $x^2 + y^2 - 2ax = 0$, with $a$ varying - i.e. circles that touch the origin).

I don't have an analytic proof for this locus, but I found it in geogebra by drawing a conic through 5 points, each of which was such an arc, and then drew a few more such arcs just to check. Most of these aren't shown in the diagram, but they are there. Assuming this elliptical locus is correct, then your point C has to lie on the ellipse in order to have $\beta=1$.

The next observation is that the point $C$ has to be the nearest point to Y lying on $p$. This is because that is the only point on $p$ such that the tangent to $p$ at $C$ will be orthogonal to $YC$, a requirement if $YC$ is to be a tangent to the circle (i.e. $\beta$ must intersect $p$ orthogonally at $C$). I should note that I realize that this argument may be flawed: If orthogonality is not actually required and $YC$ can still be tangent to the circle arc at a point on $p$, there may still be a solution. One thought I had was, similarly to the locus of arc endpoints forming an ellipse, to look at the locus of tangents from $Y$ to the family of circles whose arcs form the possible $\beta$'s. My hunch is that that locus could not possibly be exactly identical to $p$ for a portion of $p$, and then deviate elsewhere (clearly it is not the whole ellipse).

What I found when carrying out this construction was that given the point $Y$ these conditions cannot be met: If $C$ is constructed as the point on $p$ nearest to $Y$, the point of tangency to the circle arc connecting to $C$ does not lie on $p$ for the point $Y$. What I found was that the point $H$ shown, and $F$ as the nearest point on $p$, does fit both requirements.

I'm not completely satisfied with this as an answer: the elliptical locus needs to be proven (maybe to work with the coordinates you generated for $C$ and to show that they describe an ellipse), and I suppose I'd have to say it's more of an increasingly strong hunch that there is no solution as stated than a proof! I'm wondering if the length of $\beta$ is allowed to vary, if then there is a solution?

enter image description here

Here is a drawing using Narasimham's numbers to precision of 5 -- with a 7 degree discrepancy in the angle. The grey and blue points close to each other correspond to the tangent at $107^{\circ}$ and the $\beta$ endpoint at $100^{\circ}$ (precision 5 in drawing), the latter obtained by using $\frac{\beta}{r}$ radians. Narasimham's formula looks right but I'd like to be able to demonstrate it and cannot.

enter image description here

  • $\begingroup$ What is $\beta?$.. $\angle BOC$ or $ \angle AOC?$ $\endgroup$ – Narasimham May 5 '18 at 0:51
  • $\begingroup$ From the original question, I interpret $\beta$ as the circular arc length (in yellow), that is constrained to be of length 1: The angles you mentioned aren't specifically part of the definition although of course they may be relevant for a solution(?). $\endgroup$ – Circulwyrd May 5 '18 at 14:41
  • $\begingroup$ Thank you very much for your elaborate answer. I didn't understand how you construct the ellipse. Could you give more details about that? I couldn't understand the remaining of your answer because I didn't understand what $p$ is, but concerning the conclusion, I believe that there is indeed a solution. I edited my question some times ago, when I added an analytic proof that, when $\beta\in\left(0,\frac{\alpha\pi}{2}\right)$, we can find a solution. If $\alpha=2.5$, we have indeed $1\in\left(0,\frac{2.5}{2}\pi\right)$. $\endgroup$ – Scientifica May 6 '18 at 8:25
  • $\begingroup$ I used geogebra to construct points on a locus of circular arcs of length 1 which are part of the family of circles $x^2 + y^2 -2ax = 0$. The way I did it was to use the fact that the angle of the arc is $\frac{1}{r}$ in radians, so I rotated the origin point through that arc angle about points (-r,0) for several values of $r$. Then I used geogebra's 5 point conic tool on 5 of these points and it drew the ellipse $p$. I'm happy to send the geogebra file if you'd like to see it! I haven't gone through your solution in detail yet -- but I'll take a look at it this week sometime. $\endgroup$ – Circulwyrd May 7 '18 at 0:45
  • $\begingroup$ Just to further clarify, my statement in the previous comment that the angle of the arc is $\frac{1}{r}$ radians is obtained by solving the path integral from 0 to x of the parametrized circle $(r\cdot cos(t), r \cdot sin(t))$ where you know the integral is equal to $1$ (so $\beta = 1$) and you are then solving for the angle $x$. There's a little more detail to how this is carried out that I don't want to try to put in a comment, but if this part isn't clear I can modify my answer so it hopefully is better - let me know. $\endgroup$ – Circulwyrd May 7 '18 at 0:52

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