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.
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.