The blog of Lasitha Jayasinghe

Four Kissing Circles

January 25, 2026

Today we shall explore a more classical configuration which has been studied in various settings — most notably in terms of inversion. We consider $\omega_1$ , $\omega_2$ , $\omega_3$ and $\omega_4$ (with centers $O_1$ , $O_2$ , $O_3$ and $O_4$ ) to be four circles such that $\omega_i$ and $\omega_{i+1}$ are externally tangent for all $1 \le i \le 4$ where indices are considered $\pmod{4}$ .

We shall refer to the following diagram for notation. Since most labels are not standard, the precise definition of each point is given where it is introduced.

A summary of the key results pertaining to this configuration,

Points ,, and denote the tangency points of each pair of circles. An easy angle chase or an inversion about shows that quadrilateral is degenerate or cyclic, which is Problem 24 in Chapter 8 of Evan Chen’s EGMO and has also appeared as 2004 Indonesia MO Problem 4.
- Moreover, this implies that the length of the tangents from $LM \cap KN$ to $\omega_2$ and $\omega_4$ as well as the length of the tangents from $LK \cap MN$ to $\omega_1$ and $\omega_2$ are equal, which is the statement of 1999 Argentina TST Problem 2.
- Since the sum of opposite sides of $O_1O_2O_3O_4$ is equal to the sum of the radii of the circles, by Pitot’s theorem it follows that $O_1O_2O_3O_4$ is a tangential quadrilateral.
- Clearly, the perpendicular bisector of segment $LK$ is the internal $\angle O_4O_1O_2$ bisector and similarly, so it follows that the circumcenter $O$ of $LMNK$ coincides with the incenter of $O_1O_2O_3O_4$ .
- Congruency arguments show that $LMNK$ is a rectangle if and only if $\omega_1,\omega_3$ and $\omega_2,\omega_4$ are pairs of equiradial circles. In particular, when this happens $O_1O_2O_3O_4$ is a rhombus, and the incircle of this rhombus coincides with $(LMNK)$ . In this case, the antipodes of $L$ , $M$ , $N$ and $K$ in $\omega_1,\omega_2,\omega_3$ and $\omega_4$ (whenever they are defined) all lie on an ellipse with center $O$ .
- Strengthening this observation, $LMNK$ is a square if and only if all four circles are equiradial, which is asked in 2014 Kukin MO Grade 10 Problem 3.
For any point denote by the incenter of the triangle formed by the other three points.
- By the Incenter/Excenter lemma it follows that quadrilateral $LMI_KI_N$ (and similarly) is concyclic with circumcenter coinciding with the minor $LM$ arc midpoint in $(LMNK)$ .
- Further, an easy angle chase shows that all angles of quadrilateral $I_MI_NI_KI_L$ are right and hence a rectangle.
For any point denote by the orthocenter of the triangle formed by the other three points.
- It is not hard to see that quadrilateral $LMH_KH_N$ (and similarly) is concyclic, and that quadrilateral $H_MH_NH_KH_L$ is also concyclic.
- The quadrilaterals $H_MK_KMK$ (and similarly) are all parallelograms, which implies that the segments $H_MM$ , $H_LL$ , $H_NN$ and $H_KK$ concur, bisecting each other. This claim is in fact also true for any set of four concyclic points.