# What is the most efficient way to find the center of a drawn circle?

Let us assume that there is a circle drawn on a paper and I want to determine its center. I could do that in multiple ways like:

• Drawing 2 chords and intersect their perpendicular bisectors.
• Form an inscribed right angled triangle and get the center of its hypotenuse.
• Draw 4 perpendicular tangents and intersect their diagonals.
• And a lot more.

However, since we are measuring and constructing using some basics tools as rulers and compasses, some methods are more accurate than others, in other words, methods that require fewer measuring steps are more accurate due to error accumulations. Other factors may apply.

So my question is: What is the most efficient method to get the circle's center using fewest tools and yield minimum measurement errors?

• This is an interesting question, but it's not really a well-defined question. By "tools" do you mean physical objects, or construction steps, or...? What are you assuming about the measurement errors in the use of specific tools? – Greg Martin Apr 14 '17 at 20:04
• Fold it in half to get a diameter, then fold it in half again to get the centre. – Théophile Apr 14 '17 at 20:46
• @GregMartin By tools I meant physical tools like rulers and compasses. As for the definition of error based on tools, I am not sure how one can relate tools to errors in this case, is there is a way to do this? – Omar Ahmad Apr 15 '17 at 12:14
• @Théophile But that is assuming the circle lies in the center of the paper, what if it isn't? And a paper was just an example, let it be drawn in the ground. – Omar Ahmad Apr 15 '17 at 12:15
• @OmarAhmad That's why details are important! The assumption in the problem is that it's on paper. You can cut it out first before folding it. :) – Théophile Apr 15 '17 at 13:30

The reason is that in this case, missing at most by $\pm \varepsilon$ the perpendicular bissector of both the first and the second chord will generate an uncertainty square of side $2 \varepsilon$, whereas if, say, the chords make an angle equal to, $\pi/6$, the uncertainty area, instead of being a square, is a parallelogram (see picture below where the theoretical line bissectors are in red, with a margin of uncertainty $\pm \varepsilon$). The maximal length in this paralelogram is equal to $\tfrac{2 \varepsilon}{sin(\pi/12)}\approx 2 \varepsilon \times 4$: the uncertainty has been multiplied by 4 !