1 Introduction
A collection of circles in the plane, is called an arrangement of orthogonal circles if any two intersecting circles intersect orthogonally. Here, we call an intersection orthogonal, if the tangents at the intersection point form an angle of . By definition circles cannot touch in an arrangement of orthogonal circles.
A natural object that arises from an arrangement of orthogonal circles is its intersection graph. A graph is a (geometric) intersection graph if its vertices can be realized by a set of geometric objects, such that two objects intersect if and only if their corresponding vertices form an edge in . Thus, for an arrangement of orthogonal circles we define its intersection graph as the graph, whose vertices correspond to the circles in and two vertices are adjacent, if and only if the associated circles intersect in . The graph is called an orthogonal circle intersection graph.
Arrangements of orthogonal circles and their intersection graphs were recently introduced by Chaplick et al. [cfkw19]. Here it was shown that the intersection graph of circles contains at most edges. Furthermore, it is NPhard to test whether a graph is an orthogonal unit circle intersection graph. Chaplick et al. also provide bounds for the maximal number of digonal, triangular and quadrilateral cells in arrangements of orthogonal circles.
Related work.
General (nonorthogonal) arrangements of circles or disks have been studied extensively before. Giving a complete overview over the results in this field is out of scope for this article. We will hence only mention a few selected results. For the special case where all circles have the same radius the intersection graphs are known as unit disk graphs. For general arrangements of circles or balls the recognition problems for the corresponding intersection graphs are usually hard (for example for unit disk graphs [bk98]). We refer the reader to the survey of Hlinený and Kratochvíl [hk01] for more information. Other work focused on bounding the number of small faces in arrangements of circles [alps01] or about the circleability of topologically described arrangements [km14] [fs18].
Note that we can have general circle arrangements in which all circles pairwise intersect. Thus, the density of the intersection graph can be , although many graphs are not intersection graphs of circle arrangements [mm14] (for example every graph containing as a subgraph [hk01]). Hence, asking for the maximum density for intersection graphs in this setting is not an interesting question.
If the circles are allowed to only intersect pairwise in one point, then the intersection graph is called a contact graph and the corresponding arrangement is a circle packing. Due to the famous Andreev–Koebe–Thurston circle packing theorem [acpls70, Koebe36] the disk contact graphs coincide with the planar graphs. One direction of the circle packing theorem is obvious, a planar straightline drawing of the contact graph can be derived by placing the vertices at the disk centers. A related result is due to Alon et al. [alps01]. A lune is a digonal cell in an arrangement of circles. If we restrict the intersection graph of the (general) circle arrangement to intersections that are formed by lunes (we call this the lunegraph) then also in this setting we can obtain a planar straightline drawing by placing the vertices at the circle centers.
Every arrangement of orthogonal circles with the same radius can be turned into a unit circle packing by shrinking the circle size by a factor of , but there are unit disk contact graphs that are not intersection graphs of an arrangement of orthogonal circles [cfkw19].
A well established quality criteria for drawing graphs is to avoid crossings. However, crossings with large angles are considered less problematic [hhe08]. For this reason graphs that can be drawn with rightangle crossings (known as RACdrawings) are considered an interesting class from a graph drawing perspective. It was shown that graphs that have straightline RACdrawings have at most edges [del11], for being the number of vertices. Recently this approach was carried over to drawings with circular arcs that can intersect in right angles only. Chaplick et al. showed that graphs that have circular arc RACdrawings can have at most edges and there are such graphs with edges [cfkw20].
Orthogonal circle arrangements can also be seen as circular arc drawings (of 4regular graphs) with perfect angular resolution. Such drawings are known as Lombardi drawings and have been studied deeply [degkn12, e14, KKL+19].
Results.
We prove bounds for the maximal number of edges in an intersection graph of an arrangement of orthogonal circles. We show an upper bound of and present a lower bound of . As a crucial intermediate result we show that in the case of arrangements without nested circles, the intersection graph is planar. In particular, (in a similar vein to disk contact graphs and lune graphs) we obtain a planar straightline drawing by placing the vertices at the centers of the corresponding circles. As an immediate consequence we get that for arrangements of nonnested orthogonal circles the intersection graph has at most edges. We can refine the analysis to improve this bound to . This bound is tight, since we can show a matching lower bound. Our lower bound constructions can be slightly modified to also improve the bounds for the maximal number of triangular cells in arrangements of orthogonal circles to .
Organization.
2 Bounds for nonnested arrangements
For an arrangement of orthogonal circles we call the straightline drawing of its intersection graph that is obtained by placing the vertices on the corresponding circle centers the embedded intersection graph. Figure 1 depicts such an arrangement and its embedded intersection graph. In this section we prove that the embedded intersection graph is noncrossing.
We start with properties of arrangements of two or three nonnested orthogonal circles. The first observation is a simple application of Pythagoras’ theorem.
Observation 1.
Let and be two circles with centers and and radii and , respectively. Then and are orthogonal if and only if .
Lemma 1.
In an arrangement of nonnested orthogonal circles, the center of a circle is not contained inside a circle other than .
Proof.
Let and be two nonnested circles with centers and and radii and , respectively. Assume that lies inside . Obviously, and intersect, since otherwise the circles are nested. Since and intersect orthogonally it holds that . Further, since is in , we have and thus . We get that , which is a contradiction for . ∎
Lemma 2.
In an arrangement of nonnested orthogonal circles, for every pair of circles and and every point on it holds that intersects the line segment in at most one point.
Proof.
Let and be two circles with centers and and radii and , respectively. Assume that there exist a point on such that intersects twice. We call these intersection points and with and denote the midpoint between and with . By 1 lies outside of . So, for the circle to have a point inside of , has to intersect in some point (see Figure 2). Since the circles intersect orthogonally, is a right triangle. Further, since is a chord of the circle , the triangle is isosceles and its height is . Thus, we have a right angle at between and and is a right triangle.
Since the right triangles and share the same hypotenuse we get by Thales’ theorem that and have to be on the circle with diameter . We know that and . Hence, lies closer to than . It follows that is closer to than , thus . This implies that is outside of , which is a contradiction. ∎
Lemma 3.
In an arrangement of nonnested orthogonal circles, for every intersecting pair of circles and there is no third circle that intersects the line segment between the centers of and .
Proof.
Let , and be three circles with centers , and and radii , and , respectively. The circles and intersect. Assume for a contradiction that the circle intersects the line between and .
If the circle intersects the line segment between and just once, either or would be inside , a contradiction to 1. Thus, has to intersect the line segment twice. We denote these intersection points by and with and midpoint between and by . Due to 2 and cannot lie in the same circle, so one lies in and the other in . Thus, intersects both and . By 1 the center of has to be outside of the circles and . Thus, for the circle to have a point in the inside of the circles and , the circle has to intersect the circles (in some point ) and (in some point ). The situation is depicted in Figure 3.
Since all circles intersect orthogonally we have right angles at between and and at between and . Also, since is a chord of the circle , the triangle is isosceles and its height is . Thus, we have a right angle at between and . This gives us five right triangles and (red), (blue), and . We obtain
Combining these equations we get
It follows that . By a symmetric argument we see also that . We get , which is a contradiction. ∎
We can now combine our observations to prove the following result.
Theorem 1.
The embedded intersection graph of an arrangement of nonnested orthogonal circles is noncrossing.
Proof.
Suppose for contradiction four circles with centers and that are arranged in such way that their embedded intersection graph has two edges and that cross in the point . This means we have two pairs of intersecting circles and . Note that is contained in the union of and . Hence, has to lie in at least one of the circles or . By the same reasoning also has to lie in at least one of the circles or . Without loss of generality we can assume that lies in . By 1 the circle cannot enclose completely, thus it has to intersect the line segment . This, however, contradicts 3. ∎
By Theorem 1 the intersection graph is planar and we can further show that the boundary face of the embedded intersection graph is at least a pentagon if we have five or more circles (Lemma 21 in Appendix A). Applying Euler’s formula yields the following result.
Corollary 1.
The intersection graph of an arrangement of nonnested orthogonal circles has at most edges for .
3 Bounds for general orthogonal arrangements
In this section we prove an upper bound of edges for intersection graphs of orthogonal circle arrangements with nested circles. We first discuss the general approach and introduce necessary terminology before continuing with details and proofs.
For every circle in an arrangement we define its depth as the maximum cardinality of a set of pairwise nested circles in that are properly contained in . A circle with depth , i.e., it contains no circles properly, is referred to as shallow otherwise as deep.
As a first step we show that in every arrangement we can find a circle with depth at most that is orthogonal to at most seven deep circles (9). We select one circle with this property and name it the red circle. We then look at the circles properly contained in the red circle. We call these circles black circles; see Figure 4. The key observation is that we can delete the set of black circles from the arrangement and by doing so we only lose few edges from the intersection graph, i.e. at most for black circles. To obtain this bound we distinguish between intersections between the black circles and intersections between a black circle and a circle that intersects a black and the red circle (such circles are called green circles). To make our analysis work we have to partition the black circles further. If a black circle center lies on the boundary of the embedded intersection graph induced by the vertices of the black circles we call the corresponding circle boundary black circle, otherwise inner black circle.
We color edges in the intersection graph according to the color of the corresponding circles as follows: An intersection between a black and a green circle yields a green edge and an intersection between two black circles yields a black edge. If there are black circles and of those are boundary black circles, then we have at most black edges as a consequence of Euler’s formula and Theorem 1. We will prove that each black circle can be orthogonal to at most two green circles (5). However, the inner black circles can only be intersected by green circles with depth at least (11). We can chose the red circle so that there are at most seven deep green circles. The intersection graph of these seven circles has at most eight edges (2 and 12). We exploit this fact to show that only eight inner circles can be orthogonal to two green circles (13). As a final observation we show that if there are at most black circles in the red circle, there are at most inner black circles that intersect two of the green circles. We can then combine our findings to prove that we can always find a set of black circles that intersects at most circles.
We are now continuing with the proofs and details. We begin by stating a few properties of arrangements of orthogonal circles. The first lemma was proven by Chaplick et al. [cfkw19].
Lemma 4 ([cfkw19]).
No orthogonal circle intersection graph contains a or an induced .
Lemma 5.
Let and be two nested circles. There are at most two circles that intersect both and orthogonally.
Proof.
Suppose that there are two nested circles and ( lies inside ) that both intersect at least three circles , and orthogonally. Consider the intersection graph of , , , and . If the circles , and are pairwise orthogonal to each other the vertices corresponding of , , and form a , a contradiction due to 4. However, if two of the circles , and are not orthogonal to each other their corresponding vertices together with and induce a , which yields a contradiction due to 4.
Thus, at most two circles that intersect both and orthogonally. ∎
Lemma 6.
If a circle intersects the circles and orthogonally, then one of the following holds: (i) and do not intersect, or (ii) and are orthogonal and contains precisely one of the intersection points of and .
Proof.
We prove that if (i) does not hold, then (ii) holds. So assume and intersect. We apply a Möbius transformation that maps to a straight line. Note that such a transformation is conformal and thus maintains the angles; see Figure 5. The centers of and will then have to lie on . Clearly, if contains both points of then it also has to contain , but since intersects , we have a contradiction. Also, if does not contain any point of , then it has to be either contained in or is to the left or right of along , but since intersects , we have again a contradiction. ∎
Lemma 7.
In an arrangement of orthogonal circles let and be two circles that intersect. All circles that are orthogonal to and that contain the same intersection point of and are nested.
Proof.
Assume that there are two nonnested circles and that both contain the same intersection point of and . Since and contain but are not nested, they must intersect each other. Both also intersect and . This means the intersection graph of the four circles is a . This contradicts 4. ∎
The following lemma is again taken from Chaplick et al. [cfkw19, Lemma 5]. The “Moreover”part is not explicitly written down, but it is apparent from the construction given in its proof.
Lemma 8 ([cfkw19]).
Every arrangement of orthogonal circles has a circle that is orthogonal to at most seven other circles. Moreover, this circle is a shallow circle.
We can deduce a similar lemma for deep circles.
Lemma 9.
Every arrangement of orthogonal circles with nested circles has a circle with depth that is orthogonal to at most seven other circles with depth at least .
Proof.
Let be an arrangement of orthogonal circles. By deleting all shallow circles we obtain the arrangement . According to 8 we can find a shallow circle in that is orthogonal to at most seven other circles. Since is shallow in it has depth in the arrangement . ∎
In the following we select any circle that meets the requirements of Lemma 9 and refer to it as the red circle. We remind the reader that we call the circles contained in the red circle the black circles.
Lemma 10.
The set of black circles inside a red circle corresponds to a vertex set in the intersection graph incident to no more than edges, for and being the number of inner black circles in , that each are orthogonal to two circles not in .
Proof.
Let be the red circle. We count the edges incident to . Edges with two endpoints in are black edges, edges with one endpoint in are green edges. We denote the number of boundary black circles by . According to Theorem 1 the intersection graph of the arrangement restricted to the is planar. Moreover this planar graph has vertices on its outer face. Thus, by Euler’s formula we have at most than black edges.
We now count the green edges. Every circle that intersects a circle in has to intersect as well. According to 5, each of the black circles is orthogonal to at most two green circles. By our assumption black inner circles intersect at most one green circle. Thus, we have at most green edges. Adding the black edges yields the upper bound of as stated in the lemma. ∎
Lemma 11.
Every green circle intersecting an inner black circle is a deep circle.
Proof.
Let be the red circle and be the set of black circles. Suppose for a contradiction that there is a shallow green circle with center that intersects an inner circle with center . Note that in this case also has to intersect . By 1 , is therefore outside of ; see Figure 6.
Let be the arrangement consisting of the circles in and . All circles in and the circle have depth so the arrangement is nonnested. According to Theorem 1 the embedded intersection graph is noncrossing.
Let be the arrangement consisting only of the circles in . Again, all the circles are shallow, thus the arrangement is nonnested and the intersection graph noncrossing.
Since is outside it lies in the outer face of . On the other hand is an inner circle, so its corresponding vertex is not on the boundary of . The straightline edge between and must intersect an edge on the boundary of the outer face of . This yields a crossing and thus a contradiction.
∎
By 8 a red circle intersects at most deep circles. We now take a look at the possible intersections of the seven deep circles. We start with the following observation.
Observation 2.
By a case distinction we can limit the graphs that fulfil the constraints listed in 2. The proof is given in the appendix.
Lemma 12.
Every graph with at most seven vertices without an induced or has at most edges.
We can now bound the number of intersection points of the circles in .
Lemma 13.
Let be the red circle and let be the set of deep circles intersecting . The arrangement of circles in has at most sixteen intersection points of which eight are inside of .
Proof.
According to 12 the intersection graph of has at most eight edges. Hence, there are eight pairs of intersection points in the arrangement consisting of the circles in . Due to 6 for every pair exactly one intersection point is inside of . Thus, at most eight intersection points of circles in are inside the circle . ∎
Lemma 14.
In the intersection graph of every arrangement of orthogonal circles we can find a nonempty subset that is incident to at most edges, where .
Proof.
Let be an arrangement of orthogonal circles. According to 9 we can a find a red circle with depth that is orthogonal to at most seven deep circles. We denote the black circles by and set . Further let denote the vertex set corresponding to .
We now prove that there are at most inner black circles in that are orthogonal to two circles not in . According to 11 the inner black circles can only be intersected by deep green circles. If a black circle intersects two green circles, then the green circles have to intersect, otherwise the intersection graph of the black, the two green and the red circle would induce a . According to 6 a black circle that intersects two green circles contains their intersection point. 7 states that all circles containing the same intersection point must be nested. Since the black circles are not nested, only one black circle contains a given intersection point. By 13 the seven deep green circles have at most eight intersection points inside . Thus, at most eight inner black circles are orthogonal to two deep green circles. We now apply 10 with to obtain that is incident to at most edges. ∎
Our goal is to apply the last lemma for bounding the density of the intersection graph. If we can repeatedly take out vertex sets of size with incident edges (for a constant ), then the density of the graph is no more than , for being the number of vertices. Unfortunately, because of the additive constant 14 is too weak if the subsets are small. Hence, we analyse small sets separately to get a better bound. The analysis including the proofs can be found in the appendix. It culminates in the following statement.
Lemma 15.
In the intersection graph of an arrangement of orthogonal circles we can find a subset of vertices that has at most edges.
Theorem 2.
The intersection graph of an arrangement of orthogonal circles has at most edges.
Proof.
Assume there exist arrangements with orthogonal circles, whose intersection graphs have more than edges. Consider a smallest such arrangement in terms of numbers of circles and its intersection graph . By 15 there exists a subset of vertices that is incident to at most edges. We take out and all incident edges. The new graph has vertices and more than edges. This contradicts the assumption that is minimal. ∎
4 Lower bounds
In this section we discuss lower constructions. Our ideas are based on the arrangement , parametrized by two integers and , which is constructed as follows. We start with arranging circles with the same radius in such a way that their centers lie on a circle and two neighboring circles intersect. We call these circles the satellite circles. We add another circle (called hub circle) to this arrangement such that it intersects every satellite circle orthogonally. We name this arrangement a wheel of circles. An arrangement is then constructed by “nesting” wheels of circles with satellite circles each inside each other such that each satellite circle of one wheel intersects two satellite circles of the next wheel and two satellite circles of the previous wheel. The details of this construction (including the proof that the arrangement is orthogonal) can be found in Appendix D.
Lemma 16.
The intersection graph of has vertices and edges.
Proof.
The arrangement consists of wheel of circles, each having satellite circles and one hub circles. Thus, the intersection graph has vertices. Every vertex corresponding to a hub circle has clearly degree . Further, every vertex corresponding to a satellite circle has degree , except those corresponding to a satellite circle on the inner or outermost wheel of circles, which have degree . So the sum of the vertex degrees is . This number equals twice the number of edges, and therefore the intersection graph has edges. ∎
Lemma 17.
For every there is an arrangement of orthogonal circles, whose intersection graph has vertices and edges.
The proof of the lemma is obtained by counting the edges in the arrangement with and being . Details are given in Appendix B.
We now give a lower bound for nonnested orthogonal circles. The proof is given in Appendix B.
Lemma 18.
For every for which the arrangement with only the innermost hub circle is nonnested and its vertex intersection graph has edges.
Chaplick et al. [cfkw19] investigated the maximal number of triangular cells in an orthogonal circle arrangement. They proved an upper bound of and gave a lower bound of triangular cells, which they later improved to . This bound can be improved by taking the arrangement and place a small (orthogonal) circle around every intersection point. This implies the following lemma which proof is given in Appendix B.
Lemma 19.
For infinitely many values of there is an arrangement of orthogonal circles with triangular cells.
References
Appendix A Omitted material from Section 2
In Section 2 we proved that the embedded intersection graph is noncrossing. As a consequence Corollary 1 gave us an upper bound on the number of its edges. Here, we also used the fact that for orthogonal arrangements of at least five circles, the boundary face of the embedded intersection graph is at least a pentagon. We now prove this fact, which was omitted in the main part due to space constraints. We start with a helpful observation.
Lemma 20.
In an arrangement of three pairwise orthogonal circles every point in the triangle formed by the circle centers is covered by at least one circle.
Proof.
Let be three pairwise orthogonal circles with centers and radii respectively.
Suppose for contradiction a point inside the triangle not covered by any of the three circles. So its distance to any circle center is larger than the radius of that circle. Now consider the triangle , as depicted in Figure 9. We have and and since and intersect orthogonally. Combining these equations yields
It follows that the angle between and in is acute. By the same argument the angles at between and , and and are also acute. The sum of three acute angles is less than , thus we have a contradiction. ∎
Lemma 21.
The outer face of the embedded intersection graph of an arrangement of nonnested orthogonal circles is adjacent to at least vertices.
Proof.
Assume there exists a nonnested arrangement of orthogonal circles such that the outer face of its intersection graph is adjacent to exactly three vertices. It follows that there are three pairwise orthogonal circles, whose center lie on the vertices adjacent to the outer face, as seen in (a). According to 20 three pairwise orthogonal circles cover the whole triangle between their centers, thus according to 1 there cannot be another circle center within that triangle, a contradiction to .
Assume there exists a nonnested arrangement of orthogonal circles such that the outer face of its intersection graph is adjacent to exactly four vertices. By 4 these vertices cannot induce a or a , so the induced graph is a minus one edge as shown in (b). By 20 there can be no further circle inside each of those triangles (see discussion in the previous paragraph). Thus, the arrangement consists of at most circles.
As we get a contradiction in both cases it follows that the outer face of the intersection graph of an arrangement of at least circles is adjacent to at least vertices. ∎
Appendix B Omitted material from Section 3
Before we prove 12 we prove the following lemma.
Lemma 22.
Let be a graph with at most seven vertices and no induced or . Then one of the following holds:

has a vertex of degree at most .

is a , or a .

is two s glued together at a path with two edges.
Proof.
If has at most four vertices, (i) obviously holds. If has more than four vertices, we further analyse as follows.
If (i) does not hold, then contains a cycle. If the shortest cycle in has length , (ii) holds.
If the shortest cycle has length , we have two cases: If has only vertices, then (ii) holds. Or if has vertices, consider the vertex not on that cycle of length six. If has at least two neighbors, these neighbors must lie on the and thus yield a shortcut. This is not possible since the is the shortest cycle in . Hence the vertex is incident to at most one edge and (i) holds.
If the shortest cycle has length , we have three cases. First, if has only vertices, then (ii) holds. Second, if has vertices, then the sixth vertex is incident to at most one edge by the same argument used in the case of vertices and a and (i) holds. Thirdly, if has vertices and if (i) and (ii) do not hold, consider the two vertices not on the . Since (i) does not hold have degree at least two. If either one of them has two neighbors on the , it yields a shortcut and we would have a shorter circle. This contradicts the assumption that the is the shortest cycle in , so both have just one neighbor on the cycle. Since and have degree of at least they are connected by an edge. Let and be the neighbors of and on the fivecycle, respectively. The length of the shortest path from to on the is either one or two. If it is one, then the path would be a , a contradiction. So the length of the shortest path is two and is two glued together at the path from to . Thus, (iii) holds. ∎
See 12
Proof.
We delete vertices with degree until no such vertex remains. Let be the remaining graph. Since fulfils the condition (ii) or (iii) of 22 it can be made acyclic by deleting at most two edges. Hence, also can be made acyclic by deleting at most two edges. It follows that that has at most edges, if it has vertices. Thus, implies that has at most 8 vertices. ∎
In the remainder of this part we analyse arrangements with a small number of inner black circles to prove the upper bound of edges in the intersection graph of an arrangement of circles. We start with a slightly stronger statement of 14.
Corollary 2.
In the intersection graph of every arrangement of orthogonal circles we can find a nonempty subset that is incident to at most edges where . Moreover, if at most three vertices corresponding to inner black circles of are incident to two green edges each, then is incident to at most edges.
Proof.
We can follow the proof of 10 and use the fact that for the “Moreover”part we have . ∎
Before proving the desired bound in 15 we provide some necessary lemmas.
Lemma 23.
Let be the red circle in an orthogonal circle arrangement and let be the set of the black circles properly nested in . If an inner black circle of is intersected by two green circles, then contains at least boundary black circles.
Proof.
Let be an inner black circle that is intersected by the two green circles and . Since both and intersect and , they must intersect each other. Otherwise the corresponding vertices of and would induce a , which would violate 4. Thus, and intersect and according to 6 contains one of the intersection points. If is an inner circle, then its corresponding vertex in the (embedded) intersection graph of is not adjacent to the outer face. So lies on the interior of a cycle of vertices corresponding to boundary black circles. Let be the such a cycle that is inclusionminimal.
Every green circle intersecting must also intersect at least two circles corresponding to vertices in . So the vertices and corresponding to the green circles and must each have at least two neighbors in . According to 7 and since the black circles are not nested, there can only be one black circle intersecting both and . This circle is the circle . Thus, the vertices and cannot have a common neighbor.
Let be the neighbors of and be the neighbors . If two of the neighbors are adjacent, say and then induce a , which would contradict 4. Thus, muss contain at least 8 vertices. ∎
As an immediate consequence from the last lemma we get.
Lemma 24.
Let be the red circle in an orthogonal circle arrangement and let be the set of the circles properly nested in . If , then contains at most three inner black circles that each are intersected by two green circles.
Proof.
Assume for contradiction that and contains four inner circles that are intersected by two green circles each. It follows that contains at most seven boundary black circles. This is a contradiction to 23. ∎
Lemma 25.
In the intersection graph of an arrangement of orthogonal circles we can find a nonempty subset that is incident to at most edges, where . Moreover, if , then the subset is incident to at most edges.
Proof.
According to 2 there is a non empty subset in the intersection graph of every orthogonal circle arrangement that is incident to at most edges, where . However, according to 24, if then there are at most inner black circles that are orthogonal to two green circles. In this case, due to 2, is incident to at most edges.
∎
See 15
Proof.
According to 25 there is a nonempty subset in the intersection graph of every orthogonal circle arrangement that is incident to at most edges, where . Moreover, if , then the subset is incident to at most edges.
If , then is incident to at most edges.
If , then is incident to at most edges. ∎
See 17
Proof.
We set and . Note that for any positive real number we have . Hence, for our choice of parameters the arrangement has by 16
vertices. We introduce additional independent circles so that has exactly vertices. Also by Lemma 16 we get that this arrangement has at least
many edges and the statement of the lemma follows. Note that for very small our construction is degenerate, which is however covered by the bigO error term. ∎
See 18
Proof.
Let be the intersection graph of in which we have deleted all but the innermost hub circle and let its number of edges. The arrangement is nonnested by construction. Every vertex in has degree , except the vertices that correspond to the inner most satellite circles, the only hub circle and the vertices adjacent to the outer face of . The vertices corresponding to inner most satellite circles and the one hub circle have degree . The vertices on the outer face of have degree . This gives us . Thus, the intersection graph has vertices and edges. ∎
See 19
Proof.
We construct the arrangement by taking the arrangement described above and drawing a small circle over every intersection point, such that the small circle only intersects the two circles corresponding to the intersection point. That this is always possible can be seen by taking an inversion in a circle with center on the intersection point. The circles corresponding to the intersection point turn into straight lines. We can now draw a circle around their intersection point that does not intersect any other circles. Reversing the inversion gives us the small circle. Note that the inversion is a Möbius transformation and therefore conformal. Each of those new circles induces four triangular cells.
According to 16 the intersection graph of has edges. Thus, has intersection points. The arrangement has therefore circles and at least triangular cells.
We set and . Note that for any positive real number we have . Hence, for our choice of parameters the arrangement has
circles. We fill up such that it has exactly circles. So this arrangement has at least
many triangular cells and the statement of the lemma follows. ∎
Appendix C Acute nonnested circle arrangements.
We show in this part that Theorem 1 also holds in a more general setting, namely, the embedded intersection graph is also noncrossing if the intersections of circles is at an angle of at most . We call circle arrangements with this property acute (nonnested) circle arrangements. The proof of Theorem 1 relies on the combination of Lemma 1–3. We prove now these three lemmas for acute nonnested circle arrangements.
Lemma 26.
In an acute nonnested circle arrangement the center of a circle is not contained in a circle except in .
Proof.
Let and be two nonnested circles with centers and and radii and , respectively. Assume that lies inside . Obviously, and intersect in some angle , since otherwise the circles are nested. It holds that . Further, since is in , we have and thus . We get that . Since it follows that and thus . Since this is a contradiction. ∎
Lemma 27.
In an acute nonnested circle arrangement for every pair of circles and and every point on it holds that intersects in at most one point.
Proof.
Let and are two circles with centers and and radii and , respectively. Assume for a contradiction that is a point on such that intersects twice. We call these intersection points and with . The midpoint between and is denoted by .
By 26 the center of the circle has to be outside of the circle . So for the circle to have any point in the inside of the circle , the circle has to intersect the circle in a point . Since the circles and intersect in an angle , we have an angle at between and . And since is a chord of the circle , spans an isosceles triangle with height . Thus, we have a right angle at between and . It follows that
We obtain
Since it follows that . And thus, . We see that lies outside of the circle and not on , this is a contradiction. So there is no circle that intersects the the line segment twice. ∎
Lemma 28.
In an acute nonnested circle arrangement for every intersecting pair of circles and there is no third circle that intersects the line segment between the centers and .
Proof.
Let , and be three circles with centers , and and radii , and , respectively. The circles and intersect. Assume for a contradiction that the circle intersects the line segment between and .
A circle can intersect a line once or twice. If the circle intersects the line segment between and only once, either or would be inside ; a contradiction to 26. Thus, has to intersect the line segment twice. We denote these intersection points and with and the midpoint between and by . Due to 27 and cannot lie in the same circle, so one lies in and the other in . Thus, intersects both and . By 26 the center of has to be outside of the circles and . So for the circle to have a point in the inside of the circles and , the circle has to intersect the circles (in some point ) and (in some point ). This is the same situation as in Lemma 3, which is depicted in Figure 3.
Since is a chord of the circle , spans an isosceles triangle with height . Thus, we have a right angle at between and . According to the law of cosines we obtain for :
Combining these equations yields
Since it follows that and therefore . By a symmetric argument we see also that . We get , which is a contradiction. ∎
We can now combine our observations to prove the following result. The proof is identical to the proof of Theorem 1 with the strengthened lemmas. We repeat it here for completeness.
Theorem 3.
The embedded intersection graph of an acute nonnested circle arrangement is noncrossing.
Proof.
Suppose for contradiction that the embedded intersection graph has two edges and that cross in the point . This means we have two pairs of intersecting circles and , with corresponding centers . Note that is contained in the union of the convex hulls of and . Hence, has to lie in at least one of the circles or . By the same reasoning also has to lie in at least one of the circles or . Without loss of generality we can assume that lies in . By 26 the circle cannot enclose completely, thus it has to intersect the line segment . This, however, contradicts 28. ∎
An immediate consequence of Theorem 3 and Euler’s formula is the following corollary.
Corollary 3.
The embedded intersection graph of acute nonnested circles has at most edges.
Appendix D Explicit construction of the lower bound examples
We give in this section a formal construction of the arrangement including a proof of its orthogonality.
We define the following constants and points in the plane for later reference.

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