It is the classical will-my-cables-fit-within-the-tube-problem which lead me to the interest of circle packing. So basically, I have 3 circles where r = 3 and 1 circle where r = 7 and I am trying to find the least r for an outer circles of these 4 smaller circles.
After a couple of hours of thinking and some sketches with a compass I am getting close to the actual result. But how can I calculate this? With what formula?
EDIT:Thanks for the great answer. And then I come to wonder. What happens if you add another of the small circles, so you have four circles with r = 3? It is very close to 11.7
$\endgroup$ 22 Answers
$\begingroup$Clearly radius 10 is not quite enough; once you have an arrangement like this, it may be possible to find the eaxct outer radius, but in any case it can be estimated fairly well. Reminds me of these,
I drew in some trial outer circles, $r = 11$ worked with room to spare, so I split the difference, $r=10.5 = 21/2$ also worked with just a little extra room.
EDIT: did it in coordinates, I thought it was going to be a degree four polynomial but there was cancellation and it became linear, the best outer radius is $$\frac{637}{61} \approx 10.4426 $$
EEDDIITT: did it over with symbols. If the larger given radius, now 7, is called $A,$ and the smaller given radius, now 3, is called $B,$ then the radius of the circumscribed circle is $$ R = \frac{A^2 (A+2B)}{A^2 + AB - B^2}. $$
I finished up the other approach. The arrangement with larger radius $A$ and three circles of smaller radius $B,$ all tangent to an outer circle of radius $R,$ gives a cubic $$ R^3 - (A+2B) R^2 + A B R + A B^2 = 0. $$ For $A=7, B=3$ this gives $R \approx 10.397547282.$ Note that the coefficient of $R^3$ is positive, when $R=0$ the result is positive, but when $R=A$ the result is negative (for $A>B>0$). So there is a negative root, an unsuitable root $0 < R < A,$ and finally the real thing when $R>A.$ Notice that, for $A=B=1,$ we get the correct $R=1+\sqrt 2,$ meaning the centers of the four small circles are on the corners of a square, and the center of the circumscribing circle is at the center of the same square, all very symmetric in that case.
It is possible for $(A,B,R)$ to come out integers, for example $(A=9,B=5,R=15)$ or $(A=32,B=11,R=44).$ These are in the infinite family $$ A = n^3 + 4 n^2 + 4 n = n (n+2)^2, B = n^2 + 3 n + 1, R = n^3 + 5 n^2 + 7 n + 2 = A + B + 1. $$
The $(9,5,15)$ arrangement is especially good for a diagram here, as there are many visible $30^\circ-60^\circ-90^\circ$ right triangles, as well as one with sides $14,11, 5 \sqrt 3$ where I drew a pale green line in pencil.
Umm. It turned out it was possible to solve the Diophantine equation for much larger values; it is obvious that $R | A B^2,$ and a little extra fiddling with unique factorization (I'm taking $\gcd(A,B)=1$) shows that $R | AB,$ so that $AB/R$ is an integer, and the equation becomes $$ R \cdot (A + 2 B - R) = (R + B) \cdot (AB/R). $$ Here are the first hundred integer solutions
A B R A+2B-R R+B AB/R 9 5 15 4 20 3 25 22 55 14 77 10 32 11 44 10 55 8 75 19 95 18 114 15 128 93 248 66 341 48 144 29 174 28 203 24 147 62 217 54 279 42 245 41 287 40 328 35 363 244 671 180 915 132 384 55 440 54 495 48 400 183 610 156 793 120 405 118 531 110 649 90 507 395 1027 270 1422 195 567 71 639 70 710 63 605 237 869 210 1106 165 784 505 1414 380 1919 280 800 89 890 88 979 80 845 404 1313 340 1717 260 847 190 1045 182 1235 154 867 847 2057 504 2904 357 1089 109 1199 108 1308 99 1183 363 1573 336 1936 273 1296 1043 2682 700 3725 504 1440 131 1572 130 1703 120 1445 906 2567 690 3473 510 1521 278 1807 270 2085 234 1536 755 2416 630 3171 480 1568 435 2030 408 2465 336 1575 596 2235 532 2831 420 1805 1253 3401 910 4654 665 1859 155 2015 154 2170 143 1936 1845 4510 1116 6355 792 2023 895 3043 770 3938 595 2205 1672 4389 1160 6061 840 2352 181 2534 180 2715 168 2400 1477 4220 1134 5697 840 2475 382 2865 374 3247 330 2527 1266 4009 1050 5275 798 2560 597 3184 570 3781 480 2592 1045 3762 920 4807 720 2601 820 3485 756 4305 612 2645 2169 5543 1440 7712 1035 2925 209 3135 208 3344 195 3179 687 3893 660 4580 561 3249 1205 4579 1080 5784 855 3584 239 3824 238 4063 224 3645 3421 8397 2090 11818 1485 3703 2248 6463 1736 8711 1288 3757 502 4267 494 4769 442 3872 1967 6182 1624 8149 1232 3971 1076 5111 1012 6187 836 4205 4188 10121 2460 14309 1740 4335 271 4607 270 4878 255 4375 2871 7975 2142 10846 1575 4693 885 5605 858 6490 741 4704 3905 9940 2574 13845 1848 4761 2233 7337 1890 9570 1449 4851 1555 6531 1430 8086 1155 5103 3590 9693 2590 13283 1890 5184 305 5490 304 5795 288 5408 3249 9386 2520 12635 1872 5415 638 6061 630 6699 570 5600 993 6620 966 7613 840 5625 2888 9025 2376 11913 1800 5733 1364 7161 1300 8525 1092 5760 2513 8616 2170 11129 1680 5808 1745 7678 1620 9423 1320 5819 2130 8165 1914 10295 1518 5887 4411 11629 3080 16040 2233 6137 341 6479 340 6820 323 6144 5707 14048 3510 19755 2496 6727 5340 13795 3612 19135 2604 6875 2807 10025 2464 12832 1925 6877 1945 8947 1820 10892 1495 7200 379 7580 378 7959 360 7200 4939 13470 3608 18409 2640 7497 790 8295 782 9085 714 7569 4510 13079 3510 17589 2610 7623 6383 16203 4186 22586 3003 7744 1227 8998 1200 10225 1056 7840 4059 12628 3330 16687 2520 7935 1684 9683 1620 11367 1380 8019 3592 12123 3080 15715 2376 8064 2155 10344 2030 12499 1680 8112 3115 11570 2772 14685 2184 8125 2634 10975 2418 13609 1950 8379 419 8799 418 9218 399 8649 5489 15469 4158 20958 3069 8993 1353 10373 1326 11726 1173 9248 7085 18530 4888 25615 3536 9251 4491 14471 3762 18962 2871 9477 3437 13257 3094 16694 2457 9583 8835 21793 5460 30628 3885 9680 461 10142 460 10603 440 10051 958 11017 950 11975 874 10240 6061 17632 4730 23693 3520 10571 5510 17081 4510 22591 3410 10625 2036 12725 1972 14761 1700 10647 10256 24999 6160 35255 4368 10816 2605 13546 2480 16151 2080 A B R A+2B-R R+B AB/R
