I'm trying to construct (or find the centres/radii) of three tangent circles inscribed within the space between three tangent circles.

The end result should be as in this picture (the blue circles): https://www.ics.uci.edu/~eppstein/junkyard/tangencies/octahedron.html

Given the radii and centres of three touching circles, can one always construct this? Is it unique? - if not then I'm aiming to maximise the area of the inscribed circles.

At the moment I've got bogged down in some dense algebra of the type: $$ \begin{cases}|c_1 - c_a| = r_1+r_a \\ |c_2-c_a| = r_2+r_a \end{cases} $$ Where $c_{1,2}$ are the centres of two outer circles, and $c_a$ is the centre of a circle tangent to the other two. This says distances between centres of circlesis equal to the sum of the radii - hence are touching externally. I could (theoretically) do this for all pairs of tangent circles, and then work out the centres and radii in terms of $c_i$, $r_i$.

My question is - is there a neater way of finding the centres and radii of the inner circles without getting lost in the algebra? Or is there a nice geometric construction (which I could then use to implement an algorithm to find the centres/radii)?

Thanks in advance! :)


1 Answer 1


Assume that $\Gamma_A,\Gamma_B,\Gamma_C$ are three externally tangent circles, at $A=\Gamma_B\cap\Gamma_C$ and so on.
By applying a circle inversion with respect to $A$, $\Gamma_B$ and $\Gamma_C$ are mapped into two parallel lines and $\Gamma_A$ is mapped into a circle tangent to both those lines. To solve the given problem in this configuration is easy, and by applying the same circle inversion as before the original problem is solved, too.

enter image description here

The green circles are $\Gamma_1,\Gamma_2,\Gamma_3$. They are mapped into the red lines and the red circle. The yellow circles give the solution to the "red" problem and the circle inversion maps them into the blue circles, solutions of the original problem.

enter image description here


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .