The Kiss Precise by Frederick Soddy
In Nature #137, 1936 For pairs of lips to kiss maybe
Four circles to the kissing come.
Expanded to spheres : (Nature #139, 1937) To spy out spherical affairs

2(k_{1}² + k_{2}² + k_{3}² + k_{4}²) = (k_{1} + k_{2} + k_{3} + k_{4})²
When one circle is around the others, its curvature is taken < 0.
When one circle is a straight line, its curvature is 0.
This formula was first established by Descartes, but brought to fame by Soddy with a poem "The Kiss Precise", and extended to spheres, and n dimensions.
It gives the radius of the 4th circle tangent to three given tangent circles :
r_{4} = r_{1}r_{2}r_{3} / ( r_{1}r_{2} + r_{1}r_{3} + r_{2}r_{3} ± 2√(r_{1}r_{2}r_{3}(r_{1} + r_{2} + r_{3})) )
(A script)
We'll give here the proof by Philip Beecroft (1842).
First of all, consider two sets of 4 circles mutually tangent :
Circles (C_{1}) to (C_{4}) with radii r_{1} to r_{4} and curvatures k_{1} to k_{4}
Circles (Γ_{1}) to (Γ_{4}) with radii R_{1} to R_{4} and curvatures m_{1} to m_{4}
(Γ_{4}) is the incircle of triangle ABC with sides a, b, c.
Considering in the same way circles (Γ_{1}) to (Γ_{3})
with centers O_{1} to O_{3} and incircle C_{4} in O_{1}O_{2}O_{2} :
k_{4}² = m_{1}m_{2} + m_{2}m_{3} + m_{1}m_{3}
As well as all formulas from swapping the indexes (considering excircles).
We get then the sum of the squares of k_{i}'s and m_{i}'s :
∑k_{i}² = 2∑m_{i}m_{j}
and ∑m_{i}² = 2 ∑k_{i}k_{j}
Calculate then the square of the sum of k_{i}'s :
(∑k_{i})² = ∑k_{i}² + 2∑k_{i}k_{j} = ∑k_{i}² + ∑m_{i}²
And similarily (∑m_{i})² = ∑m_{i}² + ∑k_{i}²
Hence (∑k_{i})² = (∑m_{i})²,
and as they are >0 :
∑k_{i} = ∑m_{i}
Calculate then the value :
(k_{1} + k_{2} + k_{3} + k_{4})(k_{1} + k_{2} + k_{3}  k_{4}) = (k_{1} + k_{2} + k_{3})²  k_{4}² =
k_{1}² + k_{2}² + k_{3}² + 2m_{4}²  k_{4}² =
∑_{2,3,4}m_{i}m_{j} + ∑_{1,3,4}m_{i}m_{j} + ∑_{1,2,4}m_{i}m_{j}  ∑_{1,2,3}m_{i}m_{j} + 2m_{4}²
= 2(m_{1}m_{4} + m_{2}m_{4} + m_{3}m_{4}) + 2m_{4}² = 2m_{4}(∑m_{i})
And, as ∑m_{i} = ∑k_{i} :
k_{1} + k_{2} + k_{3}  k_{4} = 2m_{4}
Squaring the 4 similar relations and summing up, all terms as k_{i}k_{j} cancel and remains :
∑k_{i}² = ∑m_{i}²
Hence (∑k_{i})² = ∑k_{i}² + 2∑k_{i}k_{j} = ∑k_{i}² + ∑m_{i}² = 2∑k_{i}²
That is finally the Descartes relation :
(∑k_{i})² = 2∑k_{i}²