[b]Proposition:[/b] [i]To determine four mutually tangent, mutually exterior circles.[/i]
You, unwary observer, may have stumbled into the middle of an argument. This is a special case of Apollonius' Tangency Problem. For more information about the method used in this worksheet, a complete solution is here: [url]http://www.geogebratube.org/material/show/id/34645[/url]. [list=1] [*] [b]Bounded Arcs:[/b] The limiting position of circle C occurs when its boundary passes into a tangent line to circles A, B. A common tangent to A,B, intersects midline AB at a Similarity Point ([url]http://www.geogebratube.org/material/show/id/34182[/url]). [*] [b]Point C:[/b]Given two tangent circles, the locus of the third center can be stated as: [i]Find the locus of points equidistant from two circles.[/i] With the condition that the circles be mutually exterior, and tangent. The resulting locus is one branch of a hyperbola: [url]http://www.geogebratube.org/material/show/id/27216[/url] (the orange solution). [*][b]Similarity Axis:[/b] A Similarity axis is a straight line through any two similarity points of the circles A,B,C. Only one axis satisfies the constraints of this worksheet. [*] [math] X [/math] is the Power Center of the first three circles. A construction (Monge's problem) is here: [url]http://www.geogebratube.org/material/show/id/33929[/url]. [*] [b]Points of Tangency:[/b] Following the method outlined in the Apollonius worksheet, the points of tangency of the final (fourth) circle D can be constructed as follows: For each circle, take the closest point on the Similarity Axis, and then draw the conjugate point (like this: [url]http://www.geogebratube.org/material/show/id/34578[/url]). Draw a line joining the conjugate point to the Power Center. Where this line intersects the given circle, the new circle will be tangent. [/list] Let me know if more information would be of assistance. I am always happy to clarify my assumptions, and demonstrate them with worksheets. ______________ The Tangent Circle Problem: [list] [*]1. Tangent along the rim: solve for k [*]2a. Initial position: [url]http://www.geogebratube.org/material/show/id/58360[/url] [*]2b. Tangent to equal circles: [url]http://www.geogebratube.org/material/show/id/58455[/url] [*][b]→3a. Four mutually tangent & exterior circles (Apollonius)[/b] [*]3b. Vector reduction: [url]http://www.geogebratube.org/material/show/id/58461[/url] [/list] [list] [*]Affine Transformation [url]http://www.geogebratube.org/material/show/id/58177[/url] [*]Reflection: Line about a Circle [url]http://www.geogebratube.org/material/show/id/58522[/url] [*]Reflection: Circle about a Circle [url]http://www.geogebratube.org/material/show/id/58185[/url] [*]Circle Inversion: The Metric Space [url]http://www.geogebratube.org/material/show/id/60132[/url] [/list] Solution: [list] [*]Sequences 1: Formation [url]http://www.geogebratube.org/material/show/id/58896[/url] [*]Sequence 1: Formation [url]http://www.geogebratube.org/material/show/id/59816[/url] [*]Sequence 1: Iteration 1 [url]http://www.geogebratube.org/material/show/id/59828[/url] [*]Example of equivalent projections: [url]http://www.geogebratube.org/material/show/id/65754[/url] [*]Final Diagram: [url]http://www.geogebratube.org/material/show/id/65755[/url] [/list]