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Another argument for the impossibility of circular realizations, by Helge Tverberg, uses inversive geometry to transform any three circles so that one of them becomes a line, making it easier to argue that the other two circles do not link with it to form the Borromean rings. [27] However, the Borromean rings can be realized using ellipses. [2]
For three circles denoted by C 1, C 2, and C 3, there are three pairs of circles (C 1 C 2, C 2 C 3, and C 1 C 3). Since each pair of circles has two homothetic centers, there are six homothetic centers altogether. Gaspard Monge showed in the early 19th century that these six points lie on four lines, each line having three collinear points.
Monge's theorem states that the three such points given by the three pairs of circles always lie in a straight line. In the case of two of the circles being of equal size, the two external tangent lines are parallel. In this case Monge's theorem asserts that the other two intersection points must lie on a line parallel to those two external ...
These two circles determine a pencil, meaning a line L in the P 3 of circles. If the equations of C 0 and C ∞ are f and g, respectively, then the points on L correspond to the circles whose equations are Sf + Tg, where [S : T] is a point of P 1. The points where L meets Z D are precisely the circles in the pencil that are tangent to D.
Miquel's theorem is a result in geometry, named after Auguste Miquel, [1] concerning the intersection of three circles, each drawn through one vertex of a triangle and two points on its adjacent sides. It is one of several results concerning circles in Euclidean geometry due to Miquel, whose work was published in Liouville's newly founded ...
Descartes' theorem still applies when one of the circles is replaced by a straight line of zero curvature. If one of the three circles is replaced by a straight line tangent to the remaining circles, then its curvature is zero and drops out of equation (1). For instance, if =, then equation (1) can be factorized as [31]
If the circle centers do not lie on a line, the radical axes intersect in a common point , the radical center of the three circles. The orthogonal circle centered around of two circles is orthogonal to the third circle, too (radical circle). Proof: the radical axis contains all points which have equal tangential distance to the circles ,.
Assume that center of the circle does not lie on the line. Line-circle intersection (non-collinear case) Given a circle C(r) (in black) and a line AB. We wish to construct the points of intersection, P and Q, between them (if they exist). [13] [3] Construct the point D, which is the reflection of point C across line AB. (See above.)