Suppose you're given a circle with center $O$, I'm curious, how can one construct with ruler and compass three circles inside the larger circle such that each is tangent to the larger circle as well as to the other two?


If the radius of the circle is 1, the radius of three circles that will be internally tangent is $2\sqrt{3}-3\approx 0.464$ by Soddy's formula, given in the Soddy's circles section of this. You can construct this length, then mark it off on a diameter of the circle to find one of the centers, and finish constructing the equilateral triangle.

Added: For the construction, make a right angle with 1 on one side, swing 2 as a hypotenuse, and you have $\sqrt{3}$. Then you can mark a line to get $2\sqrt{3}-3$. Draw a diameter of the circle you are given and mark off $2\sqrt{3}-3$ from the circumfrence to find the center of one circle. Construct an equilateral triangle of $2(2\sqrt{3}-3)$ bisected by the diameter and you have the three centers.

  • $\begingroup$ Thanks Ross Millikan, I think I should have been more explicit. Do you know of a ruler and compass construction? $\endgroup$ – user7440 Feb 24 '11 at 5:05
  • $\begingroup$ Just make another line and mark off two segments of $\sqrt{3}$ end to end and back up 1 three times. This is using the modern compass that allows you to transfer a length. I have read the the Greeks used a compass that collapsed when you pick it up, but there is a proof that you can still do everything you want with one of those. $\endgroup$ – Ross Millikan Feb 24 '11 at 6:23
  • $\begingroup$ Ok, so once I get a length of $2\sqrt{3}-3$, I'm thinking of inscribing an equilateral triangle in the given circle. I could then bisect each of the angles of that triangle, and then cut of points from $2\sqrt{3}-3$ from the circumference along each of those bisectors. Like you said, these will be the three centers, correct? Thank you for your time. $\endgroup$ – user7440 Feb 24 '11 at 6:34
  • $\begingroup$ No, you can't inscribe the triangle in the unit circle as it is too small. It is sized so when you draw a circle of $2\sqrt{3}-3$ around each vertex they are tangent to the outer circle. As they need to be tangent to each other, the side of the triangle (correction to above) is $2(2\sqrt{3}-3)$. You really just need to construct an equilateral triangle of this side with its center matching the circle center. $\endgroup$ – Ross Millikan Feb 24 '11 at 14:40

sequential  steps for construction

In the image above are sequential steps of how I did it (left image is the earlier steps and the right the later steps). I'll only explain the figure constructed in the center. After having divided the disk into three (left figure) by the three radius, construct an arbitrary chord S perpendicular to AB. Construct circle centered at T tangent to the given circle at U. Construct a line passing U and the point at which the perpendicular and the the circle centered at T intersects. The line should intersect at diameter AB which is the tangent. The rest is easier to follow from the figure.

I was looking for the answer for this problem in the net but to no avail. After much trial and error, got this construction. Not 100% sure though, kinda new to this non-coordinate based geometry.

  • $\begingroup$ I was trying to figure this out ALL evening and I finally gave up and found your answer. It seems to work in GeoGebra, but when I draw it out, the inside circles are just a bit too small (twice). geogebra.org/geometry/waBjJzSn $\endgroup$ – lukejanicke Feb 26 '18 at 17:14

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