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I am considering a construction model where I have a straightedge and a constant radius compass with no access to the center: think of it as a single glass that I can trace a circle with:

Literal representation of this

While this restricted compass might feel near useless, it actually has a property that a regular compass might not have: you can easily construct a circle through any two points (as long as they are not $\geq2r$ apart), and I believe this is an important property.

My question thus is: using these two rules, is it possible to pin-point the center of any circle?

Intuitively, the answer feels like a no, since there is too much freedom for the lines and circles. I could not find any literature on this, specifically constructions where the center is not given and there can only be one radius. With this much freedom, my construction might lean into the territory of Jacob Steiner's theorem, making my answer a no, but I have no formal approach to this.

Edit 1: I know that I can find the center by having two perpendicular bisectors of two chords on the circle meet at a point. I do not know how this perpendicular bisector can be constructed: neither how the perpendicular part can be created nor dividing the chord into two.

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  • $\begingroup$ You have given 3 points and a circle template, right? $\endgroup$ Commented Sep 19, 2024 at 20:35
  • $\begingroup$ Here's an answer to your question, just use your compass to create the second circle: math.stackexchange.com/a/4096078/255730 $\endgroup$ Commented Sep 19, 2024 at 21:28
  • $\begingroup$ I don't understand the problem you pose in your edit. You can pick two points on the circle that are close enough together to connect with your straightedge and bisect with your fixed radius compass. $\endgroup$ Commented Sep 19, 2024 at 22:34
  • $\begingroup$ @Intelligentipauca maybe I didn't understand the answer given there, but I failed to recreate such construction myself. My attempt, where B is the desired center to be reached and R is the constructed one. They are indeed close, but not enough. GeoGebra link $\endgroup$ Commented Sep 20, 2024 at 7:38
  • $\begingroup$ @JohnDouma how do I bisect the line? All methods I saw online involve either the center or a variable radius compass. $\endgroup$ Commented Sep 20, 2024 at 7:49

2 Answers 2

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Here's the solution proposed in this answer with some added explanations. I'm posting this as Community wiki.

The circle on the left in figure below is the given circle. Use your fixed compass to draw another circle (on the right), intersecting the given circle at $B$ and $D$.

Draw a chord $HG$ of the circle on the right, then lines $GB$, $GD$, $HB$, $HD$, intersecting again the given circle at points $F$, $I$, $J$, $E$ respectively.

Chords $EF$ and $IJ$ are both parallel to $GH$, hence $EFJI$ is an inscribed trapezoid and we can construct the perpendicular bisector of its bases as line $LK$, where $L$ is the intersecton of $EI$ with $FJ$, while $K$ is the intersecton of $EJ$ with $FI$. Line $LK$ passes through the center of the given circle.

Repeat this construction starting from another chord (not parallel to $GH$) of the circle on the right and you'll find the center of the circle on the left.

enter image description here

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I thank Intelligenti pauca for the solution, it indeed works, and for anyone wondering what it looks like fully complete: Full construction

where W is the constructed center, precisely aligning with the real center. There are probably a couple of extra lines here and there but I am glad it works out somehow. GeoGebra link for an interactive version.

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