# The number of circles that are tangent to 3 given conics

Using Macaulay2 to generate a polynomial system solved by HomotopyContinuation.jl

In the previous blog entry I mentioned the Macaulay2 interface for HomotopyContinuation.jl that is currently under development. In this blog post I want to show how useful this interface can be.

Consider the problem of computing all circles that are tangent to 3 given conics $C_1,C_2,C_3 \subset \mathbb{R}^2$. For instance, the following picture shows 14 circles that are tangent to $$C_1 = \{y=-x^2+2x+5\}, C_2 = \{y = 2x^2+5x-8\}, C_3 = \{y = 8x^2-3x-2\}.$$

Emiris and Tzoumas write that there are 184 complex circles that are tangent to 3 general conics. This means, that there are 184 complex solutions $(a_1,a_2,r)$ such that there exists some $(x,y)\in\mathbb{C}^2$ with

• $(x-a_1)^2 + (y-a_2)^2 = r^2$,

• $(x,y)\in C_i, 1\leq i\leq 3$, and

• $(\overline{x}-\overline{a_1}, \overline{y}-\overline{a_2})$ spans the normal space of $C_i$ at $(x,y)$ for $1\leq i\leq 3$.

I wish to explore the solution space. In particular, I wish to know how many real solutions are possible, because only real solutions give circles in the real plane.

To eliminate the existence quantifier in the above equations I use Macaulay2. The following M2 code creates a file circles_conics.jl.

  R = QQ[x, y, a_1, a_2,  r,  b_1..b_6]

f = (x-a_1)^2 + (y-a_2)^2 - r^2
g = b_1*x^2 + b_2 *x*y + b_3*y^2 + b_4*x + b_5*y + b_6

Jac = diff(matrix {{x, y}}, matrix {{f}, {g}})
K = ideal(f, g, det Jac)
J = eliminate({x,y}, K)

writeSys(PolySystem {J_0}, "circles_conics.jl")


In a Julia session I include this polynomial by writing

using HomotopyContinuation
include("circles_conics.jl")


Now, the session contains a variable f which is an array of length 1. The only entry is a polynomial in the variables $a_1,a_2,r, b_1,b_2,b_3,b_4,b_5,b_6$. It vanishes if and only if the circle $(x-a_1)^2 + (y-a_2)^2 = r^2$ and the conic $b_1x^2 + b_2 xy + b_3y^2 + b_4x + b_5y + b_6 = 0$ are tangent. The following code generates 3 random assignments of the $b_i$.

circle_vars = [a_1, a_2, r]
conic_vars = [b_1, b_2, b_3, b_4, b_5, b_6]
F = [f[1](circle_vars => circle_vars, conic_vars => rand(6)) for _ in 1:3]


Computing the circles that are tangent to the three conics means computing the zeros of $F$. HomotopyContinuation.jl’s solve command does this

julia> S = solve(F)
-----------------------------------------------
Paths tracked: 512
# non-singular finite solutions:  366
# singular finite solutions:  2
# solutions at infinity:  144
# failed paths:  0
Random seed used: 24622
-----------------------------------------------


Each circle actually gives 2 solutions, one with $r$ and one with $-r$. One solution was labeled singular. The number of complex solutions is $(366+2)/2=184$.

I make a random experiment by sampling 1000 instances of the above system and counting the real solutions.

number_of_real_solutions = zeros(1000)
rands = [rand(6,3) for _ in 1:1000]

for (i, X) in enumerate(rands)
F = [f[1](circle_vars => circle_vars, conic_vars => X[:,j]) for j in 1:3]
S = solve(F)
all_solutions = results(S, onlynonsingular = false)
number_of_real_solutions[i] = length(real(all_solutions))/2
end


Here is a histogram of the results.

using Plots #The Plots package must be installed for this
histogram(number_of_real_solutions, label="b_i uniform in [0,1]", bins = 184)


The rand() command samples uniformly in the interval $[0,1]$. Two alternative sampling methods are using randn() (standard normal numbers) and rand(-20:20) (integers uniformly between $-20$ and $20$). The respective histograms are shown next.

It seems that the typical configuration of three conics has only few circles tangent to them.