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Summary
| Description | Illustration of one example where Apollonius' problem has no solutions. The three given circles are red, green and blue. The green circle divides the other two; the blue circle is entirely within, and the red circle is entirely without. Therefore, any circle (such as the black one) that is tangent to both the red and blue circles must cross the green circle; hence, it cannot be tangent to the green circle. |
| Date | |
| Source | Own work |
| Author | WillowW |
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Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled GNU Free Documentation License. |
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| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 19:53, 5 May 2008 | | 216 × 369 (823 bytes) | WillowW | {{Information |Description=Illustration of one example where Apollonius' problem has no solutions. The three given circles are red, green and blue. The green circle divides the other two; the blue circle is entirely within, and the red circle is entirel |
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