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I have two points. I need to draw an arc ($<180$°) between them, and I know how long it should be, but nothing else about it.

Knowing either the radius length or the coordinates of the center point of the circle should be enough to draw it directly. I could also calculate the radius if I knew the subtended angle θ. But I don't know how to calculate either of these.

I know that somebody answered a similar question a while ago. But for some reason, the formula they give for the answer includes θ. Maybe there is an easy way to calculate it and my knowledge is just too rusty, but I can't solve my problem with this information.

I tried deriving an answer by myself, but arrived at an equation containing both an angle's cosine and the angle squared, and I don't know how to solve such an equation. Besides, there is probably an easier way than what I used.

It is OK if the answer is derived by using trigonometry, analytical geometry or anything else, as long as I can use a pocket calculator to get my radius/coordinates. But if you don't want to lose me along the way, it would be nice if you could use calculations at undergraduate level.

Here is a drawing of what I mean: enter image description here

Community
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rumtscho
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6 Answers6

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If $r$ is the radius then you have presumably found, as Isaac did, that $l=r\theta$ and $\frac{c}{2}=r \sin(\theta/2).$ You could eliminate $\theta$ to get $$2r \sin\left(\frac{l}{2r}\right) = c.$$ That you can solve for $r$ using numerical methods.

Henry
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  • Is that any easier than using numerical methods to solve for $\theta$ with $r$ eliminated (as in my answer)? – Isaac Feb 09 '11 at 00:59
  • Only that the question title asks for the arc's radius and the text has a problem with any formula involving $\theta$. I accept that $r=l/\theta$ so there is not much difference. – Henry Feb 09 '11 at 07:10
  • I think that the second relationship should be $\frac{c}{2}=r \sin(\theta/2)$ ... or am I missing something? – Tpofofn Feb 11 '11 at 03:26
  • @Commodore64: It looks as if you are correct – Henry Feb 12 '11 at 13:56
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As you said you had, I arrive at an equation that cannot be solved algebraically, $\cos\frac\theta 2=\frac{c\theta}{2l}$, so I suspect that there is no algebraic solution, only a numerical approximation.

Isaac
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The only easier thing comes when the arc length is not much greater than the length of the straight line segment from A to B, which means the angle of the circle is small. Then you can use the small angle approximation $\cos\frac{\theta}{2}=1-\frac{\theta^2}{8}=\frac{c\theta}{2l}$, which can be solved with the quadratic formula: $\theta^2+\frac{4c\theta}{l}-88=0$ $\theta=\frac{2c}{l}+\sqrt{\frac{4c^2}{l^2}+8}$ Of course, when the difference is very small, you can just draw the straight line.

Ross Millikan
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Starting from @Isaac's solution $$\cos \left(\frac{\theta }{2}\right)=\frac{c \theta }{2 l}$$ let $x=\frac{\theta }{2}$ and $k=\frac c l$ to face $$\cos(x)=k x$$

Now, using $x_0=0$ and the rigorous form of high-order iterative methods (beyond Newton, Halley and Householder) or the easy to build $[1,n]$ Padé approximants for finding the zero of function $f(x)=\cos(x)-k x$, we should find for the very first iteration $$x_{(n)}=\frac{P_{n-1}(x)}{Q_{n}(x)}$$ The table reports the polynomials for a few values of $n$. $$\left( \begin{array}{ccc} n & P & Q \\ 1 & 2 k & 2 k^2+1 \\ 2 & 2 k^2+1 & 2 k^3+2 k \\ 3 & 24 k^3+24 k & 24 k^4+36 k^2+5 \\ 4 & -24 k^4-36 k^2-5 & -24 k^5-48 k^3-16 k \\ 5 & 720 k^5+1440 k^3+480 k & 720 k^6+1800 k^4+990 k^2+61 \end{array} \right)$$ Because ot their length, I did not write the next but, for example, for $n=8$, we would have, again as an estimate, $$x_1=\frac{-40320 k^8-141120 k^6-142800 k^4-40488 k^2-1385}{-40320 k^9-161280 k^7-201600 k^5-84224 k^3-7936 k}$$ Now, trying for a few values of $k$

$$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 0.5 & 1.0298060742 & 1.0298665293 \\ 1.0 & 0.7390847061 & 0.7390851332 \\ 1.5 & 0.5635692830 & 0.5635692042 \\ 2.0 & 0.4501836149 & 0.4501836113 \\ 2.5 & 0.3725594960 & 0.3725594960 \\ 3.0 & 0.3167508288 & 0.3167508288 \\ 3.5 & 0.2749801666 & 0.2749801666 \\ 4.0 & 0.2426746806 & 0.2426746806 \\ 4.5 & 0.2170101257 & 0.2170101257 \\ 5.0 & 0.1961642812 & 0.1961642812 \\ 5.5 & 0.1789158561 & 0.1789158561 \\ 6.0 & 0.1644189383 & 0.1644189383 \\ 6.5 & 0.1520706950 & 0.1520706950 \\ 7.0 & 0.1414307614 & 0.1414307614 \\ 7.5 & 0.1321704263 & 0.1321704263 \\ 8.0 & 0.1240396181 & 0.1240396181 \\ 8.5 & 0.1168448707 & 0.1168448707 \\ 9.0 & 0.1104342591 & 0.1104342591 \\ 9.5 & 0.1046868770 & 0.1046868770 \\ 10.0 & 0.0995053427 & 0.0995053427 \end{array} \right)$$

Notice that for $k=1$, this generates the sequence $$\left\{\frac{2}{3},\frac{3}{4},\frac{48}{65},\frac{65}{88},\frac{2640}{3571},\frac {3571}{4832},\frac{270592}{366113},\frac{366113}{495360},\frac{44582400}{6032 1091},\frac{60321091}{81615872}\right\}$$ which converges to the Dottie number.

Claude Leibovici
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If you have 2 points and arc length, it seems like you should be able to find the diameter (length of the segment connecting 2 points) by using C=πd or C/2= πr. Since you know the value of C/2 (where C= circumference and C/2 is your arc length) and you can plug in π on your calculator, you can get r, which will tell you the center of the circle. Or did I miss something?

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I think that this question can be answered very simply as follows: from the equations for the segment and section of a circle, we have the chord length, $c$ and arc length, $s$

$$ c=2R\sin\frac{\theta}{2}\\ \theta=\frac{s}{R} $$

We can determine $\theta$ by iterating to find

$$\frac{c}{s}-\text{sinc}\frac{\theta}{2}=0$$

where $\text{sinc}(x)=\sin(x)/x$. And, of course, $R=s/\theta$. I have verified this numerically for random points in the plane. Not mentioned in the original question is how to find the location of the circle center. The normal distance form the center of the chord to the origin is $d=R\cos(\theta/2)$.

Cye Waldman
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