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Van Roomen solved the general problem by solving a simpler problem, that of finding the circles that are tangent to two given circles, such as C 1 and C 2. He noted that the center of a circle tangent to both given circles must lie on a hyperbola whose foci are the centers of the given circles.
A general Apollonius problem can be transformed into the simpler problem of circle tangent to one circle and two parallel lines (itself a special case of the LLC special case). To accomplish this, it suffices to scale two of the three given circles until they just touch, i.e., are tangent.
The same inversion transforms Q into itself, and (in general) the given circle C into another circle c. Thus, the problem becomes that of finding a solution line that passes through Q and is tangent to c, which was solved above; there are two such lines. Re-inversion produces the two corresponding solution circles of the original problem.
Apollonius' problem is to construct circles that are simultaneously tangent to three specified circles. The solutions to this problem are sometimes called the circles of Apollonius . The Apollonian gasket —one of the first fractals ever described—is a set of mutually tangent circles, formed by solving Apollonius' problem iteratively.
Malfatti's problem is to carve three cylinders from a triangular block of marble, using as much of the marble as possible. In 1803, Gian Francesco Malfatti conjectured that the solution would be obtained by inscribing three mutually tangent circles into the triangle (a problem that had previously been considered by Japanese mathematician Ajima Naonobu); these circles are now known as the ...
All tangent circles to the given circles can be found by varying line . Positions of the centers Circles tangent to two circles. If is the center and the radius of the circle, that is tangent to the given circles at the points ,, then:
In mathematics, a Ford circle is a circle in the Euclidean plane, in a family of circles that are all tangent to the -axis at rational points. For each rational number p / q {\displaystyle p/q} , expressed in lowest terms, there is a Ford circle whose center is at the point ( p / q , 1 / ( 2 q 2 ) ) {\displaystyle (p/q,1/(2q^{2}))} and whose ...
Because the "sweep" of the area under the involute is bounded by a tangent line (see diagram and derivation below) which is not the boundary (¯) between overlapping areas, the decomposition of the problem results in four computable areas: a half circle whose radius is the tether length (A 1); the area "swept" by the tether over an angle of 2 ...