# How to create circles and or sections of a circle when the centre is inaccessible

I am doing landscaping and some times I need to create circles or parts of circles that would put the centre of the circle in the neighbours' garden, or there are other obstructions that stop me from just sticking a peg in the ground and using a string. Without moving sheds, walls, fence panels, etc., and without needing to go in the neighbours' garden, how would I work out the circles or sections of circles?

UPDATE: K I am getting it slowly, I like the template ideas but yes they would only be good for small curves, I am also looking at this: Finding Circular Curve When Centre Is Inaccessible

I Like this idea but I have the concern that these templates would collide with house walls, boundary fences, and obstacles as the template turns.

Here is why the template idea has practicality issues.

NOTE! Templates would not have the room to turn the distance needed. Also Patios are often laid against a house back wall so there would be a solid structure making the centre and or more inaccessible.

Also here is an example path I once had to create. approx distance to cover was about 40ft. Sorry its not exact or very good but something like this, it was nearly ten years ago now.

NEW UPDATE: It would appear this question has been answered mathematically and practically with many options, It is hard to choose an actual answer as there is more than one answer mathematically and practically. Thanks to all contributions!

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Here is a suggestion (see diagram below). Suppose you want to set out a circular arc from $A$ to $C$. The point $B$ will be on the arc as long as the angle $ABC$ is fixed.

Now I'm sure you are more up on the practicalities than I am, but I guess you could saw an angle out of a piece of wood or something. If you then perhaps put pegs at $A$ and $C$, and keep moving the angle around so that the two sides always touch the pegs, then $B$ will move through various positions which are all on a circular arc.

If the angle you use is a right angle then the section you get will be a semicircle. the angle is bigger than a right angle you will get less than a semicircle, and if it's smaller than a right angle you will get more than a semicircle. If you know one point that you want the curve to go through (apart from the ends) then you could use that to find the angle you need, otherwise you would probably have to do a bit of trial and error.

Good luck!

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Yes I remember now many years ago seeing a guy using 3 pegs set out in the above diagram with string line between them and then moving one of them backwards and forwards scoring into the lawn the shape with the middle peg, The middle peg some how rolled in the string and made the exact shape he wanted. This is a pretty vague memory though. – Maths Fail Jul 10 '14 at 7:04
The gardener's ellipse ? en.wikipedia.org/wiki/Ellipse#Pins-and-string_method – Yves Daoust Jul 10 '14 at 7:14
Yes, the pins and string would give an ellipse (oval shape) not a circle, because it would mean that the total length AB plus BC would always be the same, which is not like the angle ABC being always the same. The other thing about the pins and string method is that you might find the pegs were on the wrong side of the fence. – David Jul 10 '14 at 10:18
@Ruslan, thanks for editing the figure. Can you please tell me how you did it then I can do it myself next time ;-) – David Jul 13 '14 at 23:41
@David I just downloaded it and cropped in GIMP, then uploaded updated version. – Ruslan Jul 14 '14 at 5:17

I suggest a pointwise construction. In order to construct the circle of radius $r$ around the inaccessible point $A$, maybe you have space enough to first construct the circle of radius $\frac r2$ around an accessible point $B$ that is the midpoint of a line segment $AC$, where $C$ is also accessible. Once this has been done, you can repeatedly pick a point $P$ on this circle and construct $Q$ such that $P$ is the midpoint of $CQ$.

Alternatively, assume that the boundary to theneighbour's garden is a straight line and that the reflection $O'$ of $O$ along ths line is accessible. Use peg and string with $O'$ as center, but "reflect" the string at the border line by wrapping it around a (moving) stick there; but you have to watch out that the angles at thet stick are symmetric, or that the force acting on it by pulling is orthogonal to the border line ...

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@DacidButlerUofA Thanks for adding an illustration. – Hagen von Eitzen Jul 10 '14 at 18:43
If you can't access any points at radius of r/2 from A, then you could find a point at $\frac{2}{3}$r and draw your circle of radius $\frac{r}{3}$ and then make $Q$ three times as far from C as P. In fact, you can do this for any fraction of $r$. Cool! – DavidButlerUofA Jul 10 '14 at 19:22
OK it looks good and sounds good but for me to grasp this its going to take some studying, im not quite sure of this but the diagram helps loads thanks DavidButlerUofA – Maths Fail Jul 10 '14 at 19:27

Somehow, I feel that the most easy solution has been overlooked here:

# Draw the circle somewhere else, and then move it

1. Find a material that is affordable, or that you were going to use anyway (plastic sheet)
2. Lay it out on a place where you have access to the center point
3. Use a marker (or scissors) to create the shape that you want
4. Move the solution to your garden
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This is a really good idea, I suppose it does make it hard when the curves come from huge circles but I suppose you could always go to a field and lay out a massive sheet. It a great idea any way :) – Maths Fail Jul 11 '14 at 13:03
I agree that this is a good idea, and I thought so when I read it in nickalh's answer, which was posted 16 hours earlier. – Scott Jul 11 '14 at 15:30
@MathsFail If you have a big circle you may indeed want to go to the park. Note that you will only need a sheet as big as the area that you will actually work on in your garden. -- I must admit I got a bit stuck in the language whilst reading the post of the other person with a similar idea, but in hindsight I must give credit to nickalh for presenting the concept first. – Dennis Jaheruddin Jul 11 '14 at 15:45

Given a chord (line from one side of the circle to another, not necessarily through the centre), if you draw triangles with base as the chord and apex on the edge of the circle, then the angle at the apex is always the same. In short, the angle subtended by a chord at the centre is constant.

This also works in reverse. So if you mark out points that make the right angle when joined to the ends of your chord, then they will form a circle.

The problem is that if you use string, then the length of the string changes. You need a way to keep the angle the same, rather than the length.

One way would be to cut an angle from cardboard or wood and use elastic rather than string. Then you could move your point in such a way that the two elastic strings matched the angle.

Another would need two people. You could make an angle template with pegs in it for the string to run around. Attach the string at A and let it loop round a peg at B. Get a friend to pull on the string at B while you gently pull out at the angle so that the string doesn't kink on the sides.

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and your print screen button doesn't work? – ratchet freak Jul 10 '14 at 15:35
@ratchetfreak and my computer is not connected to the internet while on the train, but my phone is! Better-quality picture inserted now – DavidButlerUofA Jul 10 '14 at 17:56
Ok i get this one now i see what your saying, I imagine its possible especially if you have a good eye but wondering how well it would actually work in the real world. If i ever try it will let you know how it panned out. – Maths Fail Jul 11 '14 at 12:59

I am not sure if this can be made precise enough, but theoretically you could make a template with a given curvature, that will fit from the outside to your circle, and then slide it along the already made part of the circle.

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I like this idea but what keeps the template on course? – Maths Fail Jul 10 '14 at 15:40
Overlap. Just like sliding a ruler along a straight line. You mark the first part of the circle, slide the arc halfway along it, line up the overlap, and mark the new half. Repeat as needed. It will be accurate enough for most gardening. – Ross Millikan Jul 10 '14 at 15:52
That is a really good idea actually I like that one, It is a practical way for what I do, yet I am wondering how big it could be pulled off? – Maths Fail Jul 10 '14 at 16:18
I was going to suggest this, but I wasn't sure how you decide the direction of the first arc. – DavidButlerUofA Jul 10 '14 at 17:58
Yes actually I see your point DavidButlerUofA, also I wouldnt know what angle to place the first piece, I suppose it would work after a few attempts going over it until A and B meet. So now after some consideration it may be a good method but perhaps a little long due to not knowing the correct position of the first piece, I am sure there can be a quicker more precise way still. – Maths Fail Jul 10 '14 at 19:23

Use a bike.

On the side, fasten the bike to a peg with a rope having the length of the radius. By letting the bike turn around the peg, you will adjust the handlebar angle. Make sure that the wheels do not slip laterally. Then block the handlebar firmly. Your bike is ready to draw circles.

Go to the first arc endpoint, $A$, and choose a starting direction. Roll the bike on a sufficient distance to see if you will pass $B$ on the left or on the right. Then try another direction, ensuring that you pass $B$ on the other side. Then try the direction in the middle...

After a few attempts, you will reach $B$.

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I am really liking this idea but i could knock up a simple timber hinged frame with 2 small trolley wheels on the bottom at each end, then it can be used on any job any size. Hmmmm got me thinking now. Thanks Yves – Maths Fail Jul 11 '14 at 9:46
Sure, any home-made "bike" can do. If you can draw the circle on the side and you know the chord length, you can as well see the deflection between the chord and the bike that is required at $A$. – Yves Daoust Jul 11 '14 at 9:53
This answer seems to be less useful than most of the others in the situation where the desired circle is fragmented into disjoint arcs, as in the OP’s example with the “existing planters”. How would you use this solution in that scenario? – Scott Jul 11 '14 at 15:28
Very true Scott – Maths Fail Jul 11 '14 at 18:10
Soon as you have the arc endpoints, you can use the bike. To find them, I'd recommend to pinpoint them on a scaled drawing, then on the real situation. – Yves Daoust Jul 13 '14 at 14:15

Maybe not the most practical if your circles are large, but you can use the inscribed angle theorem: the angle under which a chord is seen from any point on the arc is constant.

Let the chord be $AB$ and $M$ a given third point. Nail two planks together to form the angle $AMB$ and stick pegs on $A$ and $B$. By sliding the planks against the pegs, the apex $M$ describes a circle.

As a more handy variant for large circles, nail two short planks and fasten one to a peg with a rope. The angle between the planks must be such that when the rope and the first plank are aligned with the first peg, the second peg is aligned with the second plank.

By changing the length of the rope and restoring the aligment (hold the planks at $M$, keep the rope taut, and move $M$ around $A$ until $B$ is in view), you will find other points.

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(Really a comment for Dennis Jaheruddin's answer, but comments cannot have displayed pictures.)

Draw the circle somewhere else, and then move it

At Ohio State, when the Mathematics Tower and adjoining classroom building and library were built, this came up. The architect had this circular decorative masonry motif throughout.

Mostly they could just use a string, draw the required arc, then put the bricks in. But a in few cases the center of the circle is underground. How to do it? Professor Henry Glover was the Math. Dept. representative to the construction outfit during construction, so he told them just this. Maybe lay a big piece of plywood on the ground somewhere, draw the circle on it, then move it into place for the brick-layer to use as a guide.

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Thanks for the great contribution to Dennis Jaheruddin answer. Love the brickwork, I know how dull of me. But as commented on Dennis Jaheruddin answer it can become difficult when the curves come from huger circles, some times we could only be using 20% of a huge circle. None the less I like it and will definitely be using the method, perhaps with some sheets rather than timber, although I feel as though I will be cheating my self out of what I set out to do and thats get better at maths. :/ – Maths Fail Jul 11 '14 at 13:16

My suggestion:

Assume you know two endpoints of the desired arc as well as the length of the desired radius.

From the length of the chord $C$, and the radius $R$, you can compute the length $L$ of a rope going from one endpoint to the middle of the arc. And you can repeat that from the newly created points.

The formula is $$\frac LR=\sqrt{2-2\sqrt{1-\frac{C^2}{4R^2}}}$$ or, using reduced variables $l=\frac LR$ and $c=\frac CR$ $$l=\sqrt{2-2\sqrt{1-\frac{c^2}4}}.$$ You can plot this relation once for all.

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You could try the following method, providing you know where the center of the cirle is and can access it 1 time as well as have enough room to pull a string along a tangent line of the circle.

We will use the Pythagorean Theorem to define the bounds of the cirle. Then, couple that with a tangent line to map the points of the circle WITHOUT stepping onto the neighbors property.

First, let's look at the Pythagorean Theorem $a^2 + b^2 = c^2$ and how its used to define a circle with known radius.

We are assuming you know the radius, so with an arbitrary value of side $a$ we can determine the length of $b$. Ex: $r = 10, a = 5$ Google Pythagorean Theorem Calculator yields $8.66$.

Now let's apply this to your problem.

Pull a string line represented by the blue lines. Essentially, what we have done here is create a rectangle where we want to find the length of the green segment ($x$) for an arbitrary value between $0$ and $r$. Since we can calculate $b$ we can also calculate $x$ based on the laws of a rectangle (opposite sides are equal length). With that we come up with the following formula $x = r - \sqrt{r^2 - a^2}$

Let's put this formula to work using our previous example of $r = 10$. First pick some arbitrary lengths of $a$ between $0$ and $10$. 1, 3, 5, 8, 9. This yields:

• 0.05
• 0.46
• 1.34
• 4
• 5.64

Now mark them on a perpendicular line to your tangent in both directions.

And you can see a circle beginning to form. Choose even more points to fill in the gaps as needed. Repeat this method with more tangent lines to work around obstacles.

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Its going to take a little more learning for me to grasp this one. – Maths Fail Jul 11 '14 at 12:51
@MathsFail Updated with more detail and visualizations of how this is done. – Preston S Jul 11 '14 at 15:45

A. Separate a large paper entirely from the obstacles. Draw the circle as needed on the paper with the traditional rope distance from the center. Then cut, tear or fold away the portions which conflict with the obstacles? Use pencil so you can go back and forth revising your estimate to match the needs more exactly.

Or B. Is it possible use the third dimension? Start above the obstacle with a known good circle then draw the circle you need from there?

C. Use a shadow- find or create a circle of the right size, then have a 2nd person hold the object in the right place. If necessary, bring a very bright light which can create it's own artificial shadow.

I'm disappointed there's not much math in A or C, because I love to teach it & see people overcome difficulties with math. Simplicity is best though.

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As to method C: I believe that a shadow of a circle does not automatically become a circle, but it could perhaps become one if you do things right. It seems that your method A is quite similar to my solution, I would personally recommend you to use more formatting to make sure the message really stands out. – Dennis Jaheruddin Jul 11 '14 at 15:32
Ahhh, yes, C. will be distorted unless the light source is immediately overhead. And the method of putting two points, & stretching a rope between them can approximate a circle if the endpoints are close enough to the edge of the circle. – nickalh Jul 11 '14 at 21:20

Another possible solution, using the idea of the circle of Apollonius:

Setting up a template from your scale drawing:

Pick a point inside the circle (call it B) that you are able to access in real life and find the distance from B to the centre of the circle. Call this distance c.

Make a solid template (say out of wood) with the distances c and r on it (r is the radius of the circle you want to draw). They don't have to be the real-life sizes of r and c! In fact, they just need to be long enough to produce decent changes in your stringline radius later in the process. I reckon about the size of a normal piece of paper would do it -- as long as they are to scale it will work.

We're going to need two pegs: one at B and one at a point A outside the circle. The points A and B need to make a straight line that goes through the centre of the circle.

You chose the point B already, so put a peg there. Now use your template to see how many c's it takes to get from B outwards to the edge of the circle. Count the same number of r's using your template and this is where A has to be.

Finding points on the edge of the circle

We are going to find the points on the circle by using circles at both A and B. So we'll attach a stringline to both A and B. You'll need a decent amount of string attached to both A and B, because we're going to need to make the stringlines longer as we go.

Start with the radius at A being the distance from A to the edge of the circle, and the radius at B being the distance from B to the edge of the circle (the distances AX and BX in the picture). Now make the radius at A longer by r and the radius at B longer by c. This is what your rectangle template is for. Just take the length you were already using, and move along your string the right distance, based on your template -- using the longer distance for A's string and the shorter distance for B's string.

With these new stringlines, sweep out your circle from A and your circle from B until you can find where the two circles meet. This gives you a point on the circle that you want.

Now make the string at A longer by r and the string at B longer by c and find the new intersection. And again make the string at A longer by r and the string at B longer by c, and so on. This will construct a series of points on the circle which you can join up to make the circle you want. (It may be that you can't reach all of the circle using these pegs, so you might have to do this from several places.)

Why this works (for the mathsy readers)

Apollonius' definition of a circle is that it is the locus of points whose distances from two particular points is in the same ratio. One of these is inside the circle and the other is outside.

Using the naming of points in my first diagram: If you have B, you know that the ratio of AX to BX has to be the same as the ratio of AY to BY (where Y is the point on the far side of the circle from X). But BY = 2c + BX and AY = AX + 2r. So: \begin{align} \frac{AY}{BY} = \frac{AX+ 2r}{BX+2c} &= \frac{AX}{BX}\\ AX \times BX + 2r BX &= AX \times BX + AX \times 2c \\ 2r BX &= AX \times 2c\\ \frac{r}{c} &= \frac{AX}{BX} \end{align} So distances from $A$ to the circle must always be $\frac{r}{c}$ times distances from $B$ to the circle. This is a linear relationship, so if you make one distance longer by $r$ then the other has to be longer by $c$.

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Its going to take some learning for me to grasp this, but I am on it. – Maths Fail Jul 11 '14 at 12:50

One alternative is to plot several points along the circle using Cartesian coordinates. This has already been suggested, but it can be generalized to deal with many kinds of obstacles.

If you did have free access to all points in the interior of the circle, you could set up your Cartesian coordinate system by laying two straight reference lines perpendicular to each other to form two diameters of the circle, like so:

For a circle of radius $r$, you could make a table of $x$ and $y$ values using the formulas $x = r \cos \theta$ and $y = r \sin \theta$ for a sequence of angles. For each pair $(x,y)$, you measure a distance $x$ along one of your reference lines to find point $A$ and a distance $y$ along the other reference line to find point $B$. Then attach a string of length $x$ at $B$ and a string of length $y$ at $A$ and extend the two strings taut so that they meet at $C$, which is a point on the circle.

After plotting several points regularly spaced around the circle, you can use a curved template (constructed elsewhere) to mark the arc of the circle between each pair of points. The number of points you need to plot is a function of how long you can make your template relative to the radius of the circle. For example, if you can build a template that is slightly more than a $10$-degree arc of the circle, you only have to use a table where the angle $\theta$ is given in $10$-degree increments.

Now to generalize: the two reference lines do not need to be diameters. You can offset one or both of the lines from the center of the circle, and as long as they are perpendicular to each other, you merely need to add or subtract the amount of each offset from the relevant coordinate. For example, let's move one of the reference lines $a$ feet from the center and the other reference line $b$ feet from the center. The result looks like this:

Now to plot the point that was at $(x,y)$ in your original table of coordinates, you place points $A$ and $B$ at distances $|x-a|$ (or $x+a$) and $|y-b|$ (or $y+b$) along the reference lines, measured from where the lines cross, and then measure the same distances from the points $A$ and $B$ in order to find a point $C$ on the circle.

Now, for example, to deal with the shed in the middle of the circle, you lay your reference lines along two sides of the shed. The distances $a$ and $b$ in this case are just half the dimensions of the shed. You can then plot more than half the circumference of the circle. To plot the remaining part of the circle, lay reference lines along the other two sides of the shed.

For the circle cut off by the neighbor's fence, lay one reference line along the fence (or use the fence itself if it is straight enough) and lay the other reference line perpendicular to the first. In this case you only need one of the lines (the one parallel to the fence) to be offset from the center, which simplifies the task; you can plot two symmetric points using each $(x,y)$ pair.

For the patio at the edge of the pool you might find it convenient to lay the two perpendicular reference lines so that they are tangent to the circle you want to plot. That is, the offsets are $a = r$ and $b = r$.

For the planters, you could (in the example shown) lay the two reference lines just a bit off-center while staying clear of all the planters, but then you might not be able to plot as many points as you might like (because planters would interfere with the lines from $A$ or $B$ to $C$). You might find it easier to lay out a square larger than the circle, but with the same center, and use adjacent sides of that square as reference lines for the various parts of the circle. That is, you can let $a = b > r$. But in the example shown, there is one arc of the circle that cannot be plotted by any pair of sides of the square, so you would have to lay another reference line between the planters, perpendicular to the nearby side of the square, in order to plot that arc.

By the way, for a circular curved driveway of uniform width, if you make templates for both the inner and outer radii of the driveway and attach them to a rigid frame so that their arcs are concentric (perhaps by attaching plywood "blanks" to the frame, drawing the arcs on them from a common center point, and then cutting the arcs), you need only plot points along one edge of the driveway; every time you use the template to fill in the points along that arc, you can use the other side of the template to fill in points along the other arc of the driveway.

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Here's another option that should be easy to implement and hopefully is exact enough. You can approximate the arc of a circle with an ellipse by adjusting the distance between the foci of the ellipse depending on how close you can get to the center. Once you've found the foci, you drive your stakes there, tie off the right length of rope across them and scribe the arc.

First, the math: Start with a circle at the origin. For a fixed y-distance from the center of the circle (written as $pR$ where $p$ is the percentage of the Radius), I center an ellipse so that the major axis is the chord of the circle through that closest point of approach to the center of the circle, perpendicular to the y-axis. (This puts the ends of my ellipse on the circle.) I set the minor axis so that the top of the ellipse is also on the circle. That defines a unique ellipse.

In the diagram, the target circle of radius $R$ is blue, the constructed ellipse is green. $X$ is the center of my ellipse. $B$ is the length of the semi-major axis. $A$ is the semi-minor. $C$ is the distance to the foci. $Y$ is the center of my desired arc.

(Note on choices: You could get a better fit with different methods, such as splines, or possibly by using different fixed points to construct the ellipse, but I chose this method to make the actual measurements and construction in the field as dirt-simple as I could, assuming that the errors incurred from cutting and tying string, walking around, and measuring across the lawn would out-weigh any gains I would get by complicating it.)

Now how do you implement it. The good news is, all you need to compute is $C$. Everything else you can measure. I'll do this assuming a 100 foot radius circle, and then all of our units will be in feet and can be compared as a % of the radius. Everything scales, so if we're within 1 foot of the circle for a 100 foot radius, then we're within 1%, and we'll be within 2 feet on a 200 foot one, $1\over2$ a foot on a 50 foot one, etc.

Start by driving a stake in the center of your arc (point $Y$ on the diagram), then find the center line to the middle of your circle. Find the closest point ($X$ on the diagram) you can get to the center of the circle so that you are unobstructed on either side of the line (your left and right as you face the center), and write the distance to the center of the circle as a percentage of the circle's radius. So, if you're 10 feet from the center of our 100 foot circle, that's $10\%$, or $p=0.10$. Now compute $C$ as

$C = \sqrt{2pR^2(1-p)}$,

then drive a stake at a distance $C$ on either side of the center line (point $X$); these are your ellipse foci. Tie a string around one of the foci stakes, loop it around the stake at the center of your arc (point $Y$), then tie the other end to the other foci stake. Pull up the center stake from your arc and use the loop to scribe the arc.

According to the numbers I ran, the accuracy starts to drop off a bit once the width of the arc that is about the same as the circle radius, that's why I drew a chord across the top of the diagram with a length of $R$. It would take 6 of these arc sections to completely draw the circle. Across this arc, your % error is about ${1\over10}p$. So in our example, where $p=10\%$, your error is about $1\%$, or within 1 foot. And everything scales pretty well, so at 20 feet out your error is less than $2\%$, or 2 feet for the 100 foot case, etc. It tapers off further out, so even out at $70$ or $80\%$ you're still under $5\%$ error.

Hopefully, this level of error is ok, especially given that you only need to make 3 measurements and drive 3 stakes to whip this out. (Maybe 4 stakes, adding one at $X$ and running a string between $X$ and $Y$, and another between the foci to ensure they are perpendicular.)

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Ok going to have to do a little learning for this one but interesting. – Maths Fail Jul 11 '14 at 12:53

Hire some aliens to do a crop circle for you? :)

If you have a good "baseline", such as the fence, you want to work from, I would think that you could (with paper and pencil) treat the fence as a chord within the circle, and figure out a perpendicular to this chord every few feet along the fence, and the distance to the circle along each perpendicular. You don't get a continuous arc, but should have points close enough that you can eyeball the arc between them.

For the planters scattered throughout a circular patio, you could use the sides of various planters as your chord (extend it to the circle with a length of lumber or something) and do the chord perpendicular trick in a piecewise fashion.

Sorry for no pictures, but this box is too small to contain the proof... :)

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Let say $A,B$ is 2 points on the circle you want, and the centre $O$ is inaccessible. Well simply flip $O$ over $AB$, draw circle around the new centre (which is no longer inaccessible) to get an arc from $A$ to $B$, then flip that arc over $AB$ to get the arc of the original circle.

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One more option, inspired by Hagen von Eitzen's answer, but allowing you to construct the circle from the outside:

Step 1:

Pick two points that you know your cricle is going to pass through. Perhaps where the circle meets the fence, for example. I'll call these points A and B.

Step 2:

Construct a "scale drawing" of your situation on the ground somewhere not too far away.

In essence, pick a point and use a stringline to draw a circle. Then mark on your circle matching points to A and B. You could, if you want actually draw a scale drawing on a really big piece of paper and lay it on the ground.

I'll call the drawing of A the point A', and the drawing of B the point B'.

It is important that your new line A'B' is parallel to the real AB. It is also important that the circle in your drawing is smaller than the real circle needs to be.

Step 3:

Draw lines from the real A to A', and from the real B to B', and find where these lines meet. Call this point C.

Now we're all set up.

Step 4:

Get a long piece of elastic and attach it to a peg at C, then stretch it out from C to A (or C to B if it's longer), Make sure it's well stretched when the end is at A (or B).

While it's stretched out, it should touch the drawing of A. Mark this point of the elastic in some way (perhaps tie a coloured cloth to it at this spot).

Step 5:

If you move the end of the elastic in such a way as your coloured mark follows the small circle, then the end of the elastic will trace out the big circle.

It will be tricky to make your line smooth, so probably marking out particular points and then filling it in between by eye afterwards is probably the best plan.

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