In \(D\)-dimensional euclidean space, the quadrance of a vector \(X\) is non-negative:
$$ \Vert X \Vert^{2} \geq 0. $$
Furthermore, the inner product between two vectors \(X\) and \(Y\) satisfies
$$ -1 \leq \frac{X \cdot Y}{\Vert X \Vert \Vert Y \Vert} \leq 1. $$
Consider the position vectors for 4 points in \(D\)-dimensional euclidean space:
$$ X_{1}, \qquad X_{2}, \qquad X_{3}, \qquad X_{4}. $$
With 4 vectors you can construct 6 vector differences \(X_{ij} \equiv X_{i} - X_{j}\):
$$ X_{12}, \qquad X_{14}, \qquad X_{32}, \qquad X_{34}, \qquad X_{42}, \qquad X_{13}. $$
However, only 3 of these differences are linearly independent. For example,
$$ X_{34} = X_{14} + X_{32} - X_{12}, \qquad X_{42} = X_{12} - X_{14}, \qquad X_{13} = X_{12} - X_{32}. $$
You can introduce the following quantities:
$$ p \equiv \frac{\Vert X_{14} \Vert}{\Vert X_{12} \Vert}, \qquad x \equiv \frac{X_{12} \cdot X_{14}}{\Vert X_{12} \Vert \Vert X_{14} \Vert}, \qquad a \equiv \frac{X_{14} \cdot X_{32}}{\Vert X_{14} \Vert \Vert X_{32} \Vert}, \qquad y \equiv \frac{X_{12} \cdot X_{32}}{\Vert X_{12} \Vert \Vert X_{32} \Vert}, \qquad q \equiv \frac{\Vert X_{32} \Vert}{\Vert X_{12} \Vert}.$$
The quadrance of \(X_{34}\) can be written as
$$ \Vert X_{34} \Vert^{2} = \Vert X_{14} \Vert^{2} + \Vert X_{32} \Vert^{2} + \Vert X_{12} \Vert^{2} + 2 (X_{14} \cdot X_{32}) - 2 (X_{14} \cdot X_{12}) - 2 (X_{32} \cdot X_{12}). $$
Thus,
$$ \frac{\Vert X_{34} \Vert^{2}}{\Vert X_{12} \Vert^{2}} = 1 - 2xp -2yq + p^{2} + 2apq + q^{2}. $$
In a similar fashion, you also have
$$ \frac{\Vert X_{42} \Vert^{2}}{\Vert X_{12} \Vert^{2}} = 1 - 2xp + p^{2},$$
and
$$ \frac{\Vert X_{13} \Vert^{2}}{\Vert X_{12} \Vert^{2}} = 1 - 2yq + q^{2}.$$
With 4 position vectors, you can construct two distinct conformal ratios:
$$ u \equiv \frac{\Vert X_{12} \Vert^{2} \Vert X_{34} \Vert^{2}}{\Vert X_{14} \Vert^{2} \Vert X_{32} \Vert^{2}}, \qquad v \equiv \frac{\Vert X_{13} \Vert^{2} \Vert X_{42} \Vert^{2}}{\Vert X_{14} \Vert^{2} \Vert X_{32} \Vert^{2}}. $$
In terms of the polynomials from above, you have
$$ u = \frac{1}{p^{2} q^{2}}(1 - 2xp -2yq + p^{2} + 2apq + q^{2}), $$
and
$$ v = \frac{1}{p^{2} q^{2}}(1 - 2xp + p^{2})(1 - 2yq + q^{2}). $$
Note that for \(\lambda \neq 0, -1, -2, ...\) you have
$$ \left( \frac{1}{v} \right)^{\lambda} = p^{2\lambda} q^{2\lambda} \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} p^{m} q^{n} C_{m}{}^{\lambda}(x) C_{n}{}^{\lambda}(y). $$
Here \(C_{n}{}^{\lambda}(x)\) is a univariate Gegenbauer polynomial. On the other hand, for \(u\) you can write
$$ \left( \frac{1}{u} \right)^{\lambda} = p^{2\lambda} q^{2\lambda} \sum_{m = 0}^{\infty} \sum_{n = 0}^{\infty} p^{m} q^{n} C_{mn}{}^{\lambda}(x,y \vert a). $$
Here \(C_{mn}{}^{\lambda}(x, y \vert a)\) is a bivariate Gegenbauer polynomial.
