the chain is from the yellow circles.

What I want is a way to draw the next yellow circle given all the ones before it:

enter image description here

I know how to draw the first circle $(P_0Q_0X)$.

I know the (nameless in the picture) contact point of the $n-th$ circle with the $n+1-th$ one lies on a circle centered at $H = P_0Q_0 \cap BC$ passing through $C$.

I know lines $P_nQ_n$ all meet in $H$.

I know quads $P_nP_{n+1}Q_{n+1}Q_n$ are cyclic.

I still can't find a simple way to construct the next circle given the previous one.I know there are one or two inversions that can do the trick but I would prefer if we avoid the temptation of looking for invertions.

I can't prove but I know the circles $P_nQ_nC$ are tangent to $BC$ at $C$

EDIT: also don't just use the general apollonius solution unless you can make sure you can show some symetry from this particular problem. For example: because we know the contact points between two circles lie on a circle centered at $H$ passing through $C$ we don't need the full $CCC$ but we can use $PCC$ (of course you guys are suposed to show more simplifications)

  • $\begingroup$ If you agree to use conics: the center of each circle in the chain lies on an ellipse with foci L and B. Also, $Q_{n+1}$ lies on hyperbola with foci L and $Q_n.$ $\endgroup$
    – user376343
    Aug 1, 2020 at 20:01
  • $\begingroup$ thanks but i rather to use the straight edge and compass $\endgroup$ Aug 1, 2020 at 23:01
  • $\begingroup$ I know there are one or two inversions that can do the trick ... You should describe those inversion arguments (or any other arguments you know). Someone may be able to use them as jumping-off points for a non-inversive approach, without wasting time re-deriving key metric relations and such. $\endgroup$
    – Blue
    Aug 3, 2020 at 20:23
  • $\begingroup$ the main inversions I want to avoid are the ones centered at $C$. One very interesting inversion that I would accept is the one around the circles $(P_nQ_nC)$ $\endgroup$ Aug 3, 2020 at 20:41

2 Answers 2


CircleChain We use the point $H$ and the circle that passes through all the contact points of neighboring circles of the chain, which was mentioned by the OP in his post. By the way, $H$ is located on the segment $LB$ a distance of $\frac{a}{6}$ away from the point $L$.

Start the construction by drawing the aforementioned circle, which is marked by the two end-points $C$ and $D$ and has its center located at $H$. Furthermore, its radius is equal to $HM_{01}$. Remember that the point $M_{01}$ is already available at this moment. This circle cuts the circle $O_1$ at $M_{12}$. Draw and extend the line $O_1 M_{12}$. We know that the center of the sought circle of the chain lies on this line.

Now, draw the two lines $M_{12}H$ and $O_1L$. The line $O_1L$ passes through the contact point $Q_1$ of the blue and yellow circles. Then, draw a line perpendicular to the $O_1L$ at $Q_1$ to intersect $M_{12}H$ at $N$. Construct an auxiliary circle with radius $Q_1N$ and having its center at $N$. This circle meets the blue circle at $Q_2$. The line $Q_2N$ is the common tangent of the blue circle and the sought member of the circle chain. Furthermore, $Q_2$ is the contact point of these two circles. Therefore, the center of the sought circle lies on the extended portion of the line $Q_2L$.

Now, we have two lines harboring the center of the sought circle, i.e. $O_1 M_{12}$ and $Q_2L$. Therefore, the point $O_2$, where these two lines meet, is the center of the next member of the circle chain. To complete the construction, draw the circle with radius $O_2Q_2$ or $O_2M_{12}$ taking $O_2$ as its center.

$\underline{\mathrm{Added\space at\space OP’s\space Request\space …}}$

A geometric configuration, in which three circles (e.g. green, yellow and blue circles) touch each other externally, has a unique point (in our case $N$), where the three common tangents coincide. Therefore, we can draw any two of the three common tangents to obtain this point. In other words, the point of intersection of any two common tangents (e.g. $NM_{12}$ and $NQ_1$) gives us this point. That is how we obtained the point $N$ in the first place. Using $Euclid\space Theorem\space 59$, we can show that the three distance from this point to each contact point of a pair of circles are equal. That is why we constructed an auxiliary circle with radius $Q_1N$ (or $NM_{12}$) and having its center at $N$. This circle cuts the blue circle at $Q_2$ giving us the third common tangent $NQ_2$. Now, we know that there exist a unique circle, which touches the yellow circle at $M_{12}$ and the blue circle at $Q_2$ externally. The center of this circle lies at the point of intersection of the two lines $O_1 M_{12}$ and $Q_2L$. The radius of this green circle is chosen as $O_2 Q_2$ to let its circumference pass through the point $Q_2$.

However, the story does not end here, because we have not yet explicitly stated that the green circle we obtained touches the red circle internally. This can be proven using trigonometry. But first, we would like to put forward the following argument. If the green circle cuts or fail to touch the red circle, then, this problem has no solution, because you cannot find another circle that touches the red circle internally and the blue circles externally while touching the yellow circle at $M_{12}$ externally.

If you want us to post the proof, please let us know

  • $\begingroup$ thank you very much, but why is point $Q_2$, as drawn, on the green circle? I do understand that $Q_2L$ is tangent to the auxiliary circle. But how do we know $Q_2$ is on the green one? $\endgroup$ Aug 7, 2020 at 23:57
  • $\begingroup$ I see it now it is because $HM_{12}$ is radical axis $\endgroup$ Aug 8, 2020 at 0:15
  • $\begingroup$ @hellofriends I just added two paragraphs to my answer to clear the doubts you expressed in your first comment. If you want to know the proof, please let me know. Thanks for the bonus $\endgroup$
    – YNK
    Aug 8, 2020 at 16:57

enter image description here

The diagram is copied here for convenience. I'll refer to the circle with center $L$ as the blue circle and the circle with center $B$ as the white circle.

We're trying to find the blank yellow circle, given the previous circle $P_3Q_3C_3$, where $C_3$ is the unnamed contact point.

Construction using inversion: For the circle $c$ through the points $P_n,Q_n,C$, we can invert $P_{n-1}$ and $Q_{n-1}$ in $c$ to get $P_{n+1}$ and $Q_{n+1}$. You have also mentioned that you can construct the contact points. So once the first two circles have been constructed we can leapfrog to construct the rest of them.

To show this, invert the diagram in a circle with center $C$.

Construction not using inversion: Let $L'$ be the point on the upper half of the blue circle such that $L'L$ is perpendicular to $BO_3$. Then let $Q_4$ be the other intersection of $L'P_3$ with the blue circle.

Similarly, let $B'$ be the point on the lower half of the white circle such that $B'B$ is perpendicular to $LO_3$. Then let $P_4$ be the other intersection of $B'Q_3$ with the white circle.

We now have 3 points of the blank yellow circle, so we can construct the circle and its center.

Note 1: this is an adaptation of a construction at Eppstein's The Geometry Junkyard and is really just a $PCC$ Apollonian construction. The problem is also equivalent to finding the incenter of a hyperbolic triangle. Baragar and Kontorovich's Efficiently constructing tangent circles claims to have an even shorter construction but I have not tried it out.

Note 2: You can prove that circles $P_nQ_nC$ are tangent to $BC$ by inverting in a circle centered at $C$. They map to vertical lines.

  • $\begingroup$ that's very good. I don't think anyone will come with a more better solution. In a couple days I will give you the full prize. Thanks! $\endgroup$ Aug 6, 2020 at 2:48
  • $\begingroup$ i believe there is a typo, you meant $B'Q_3$. Then again I wrote "more better" $\endgroup$ Aug 6, 2020 at 3:40
  • $\begingroup$ Yes, thanks @hellofriends. Fixed. $\endgroup$
    – brainjam
    Aug 6, 2020 at 4:29
  • $\begingroup$ your answer is very rich but I feel like the other one used the simetries a bit better and it would be unfair to give you the bounty when the other managed to simplify a lot the problem. I will accept yours and give the points to the other guy. $\endgroup$ Aug 8, 2020 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.