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Is there an approach to this problem that uses the Theorem of Descartes circles? or some other slightly general method? I have looked at scores of links about these topics on this site and elsewhere, and can use Descartes' curvature theorem for easier problems where we are given 3 tangent circles, find the 4th. I don't know how I could apply it here and I didn't really follow one of the articles about extending it to more circles. Extension of Descartes' "Kissing Circles" Theorem Maybe it is just that my circles are not mutually tangent?) I would also like to be able to toss in another tangent circle somewhere and still be able to find the surrounding tangent circle. I wouldn't be able to do that by the symmetry method that I have used.

6 plus 1 circles

Problem Statement: Each of 6 circles is radius 1, and the bottom row of 3 circles have the same y-coordinate. Two identicle circles are closest packed right on top of the first row. A third identicle circle finishes the stack on top. Find the equation of the surrounding circle that is just tangent to the three corner circles. See figure.

I was successful by the following method. 1) Observing symmetry, I could calculate the point coordinates of J. 2) I let point G(x,y) and H(x,y) be unknowns. 3) I set up 4 equations and solved them simultaneously.

Eq1: |J−(Gx,Gy)|=|J−(Hx,Hy)||J−(Gx,Gy)|=|J−(Hx,Hy)| Equilateral triangle legs

Eq2: |J−(Gx,Gy)|=|(Gx,Gy)−(Hx,Hy)||J−(Gx,Gy)|=|(Gx,Gy)−(Hx,Hy)| Equilateral triangle legs

Eq3: (Gx−1)2+(Gy−1)2=1(Gx−1)2+(Gy−1)2=1 Point G has to be in the lower left circle

Eq4: (Hx−5)2+(Hy−1)2=1(Hx−5)2+(Hy−1)2=1 Point H has to be in the lower right circle.

Once I had point J, G, and H, finding the circle center and radius was easy enough.

Aside: Tangent circles, in close pack configuration, has interested me for a long time because it is relevant to pharmaceutical vials in various trays and loading equipment.

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  • $\begingroup$ What is your question? It looks like you have solved it... $\endgroup$ – QuIcKmAtHs Jan 20 '18 at 1:44
  • $\begingroup$ Absolutely not solved! The question is what is another method? The problem is just 1 example of a class of problems, most of which can't be solved the way this one was. I just didn't want someone showing me the method I already knew, which by and by won't work most of the time. $\endgroup$ – Narlin Jan 20 '18 at 1:53
  • $\begingroup$ Well u want a solution containing analytic geometry or a pure geometry? $\endgroup$ – Pookaros Jan 20 '18 at 3:35
  • $\begingroup$ The difficulty is that you specialize a broad class of minimum bounding circle problems to one involving "tangent circles" to some extent, and then want to be told something you don't already know. Did you know the number of points of contact between the minimum bounding circle and the circles being enclosed is either two or three? If you can identify those two or three points, they define your bounding circle. $\endgroup$ – hardmath Jan 20 '18 at 4:59
  • $\begingroup$ @hardmath I did not know the number of points of contact between the minimum bounding circle and the circles being enclosed is either two or three? and while I am certainly going to look into that and hope to convince myself, it isn't right off obvious too me. thank you. $\endgroup$ – Narlin Jan 20 '18 at 17:39
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What I would do for a general solution Let's name the circles with radius r $C_{1-6}$ from bottom left to top right and $C$ the big one.

I would start from $C_2 , C_4, C_5$ taking their centers and forming a triangle $T_1$ with them. This triangle obviously would be isosceles with sides 2r. It's centroid is is point $M$ in the figure (because the lines defining it intersects 2 circles at the time, so they divide the triangles sides in 2 equal pieces) and its also the center of the large circle$^*$. We know that since its a centroid it's distance $d$ from the side of the triangle is 1/3 of it's height $h_1$, so with a pythagorean $h_1 = \sqrt{r^2 - (\frac{r}{2} )^2}=\sqrt{\frac{3}{4}r^2}=\frac{\sqrt 3}{2}r$ so $d=\frac{\sqrt 3}{6}r$.

Now working with $C_4,C_5,C_6$, we define the triangle $T_2$ and its height $h_2=h_1 $since $T_1 \equiv T_2$. Last the radius of $C_6$ is also r.

Now notice that $R=d + h_1 + r=\frac{\sqrt3}{6}r^ + \frac{3\sqrt3}{6}r^ +r=\frac{2\sqrt 3 + 3}{3}r$ where $R$ the radius of the big circle and so we have the radius.

The center $M$ is easy to find since we know its distance from known circle centers. That's all u need to define $C$'s equation

$^*$I need to also prove that $M$ is a center as lema. If the above are true then its the only thing missing for the proof to be full

P.S. Also yes this does not correspond to the kissing circles theorem cause thy are not mutually tangent. These has relation to Apollonian circles. https://sites.math.washington.edu/~julia/teaching/445_Spring2013/DescartesAndTheApollonianGasket%20(1).pdf

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  • $\begingroup$ This answer is useful as it generalizes a method that works for symmetric tangent circles as shown in the figure and the number of circles can be extended as large as needed. Also, it is different and easier to think about than the solution shown in the problem. It hasn't been marked as accepted, only because it doesn't generalize to a set of asymmetric tangent circles. $\endgroup$ – Narlin Jan 20 '18 at 18:43

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