Given $x_1,x_2,y\in\mathbb R^d$, I want to write $$\frac{\|x_1-y\|^2}{\sigma_1^2}+\frac{\|x_2-y\|^2}{\sigma_2^2}\tag1$$ in the form $$\left\|\frac{x_1}{r_1}-\frac{x_2}{r_2}\right\|^2+b\|y-z\|^2\tag2$$ for some $r_i,b\in\mathbb R$ and $z\in\mathbb R^d$ independent of $y$.
I've tried a lot and I'm not able to figure out how I need to choose them.
Note that $(1)$ expands to $$\frac1{\sigma^2_1}\|x_1\|^2+\frac1{\sigma_2^2}\|x_2\|^2+\left(\frac1{\sigma_1^2}+\frac1{\sigma_2^2}\right)\|y\|^2-2\left\langle\frac{x_1}{\sigma_1^2}+\frac{x_2}{\sigma_2^2},y\right\rangle\tag3$$ and $(2)$ expands to $$\frac1{r_1^2}\|x_1\|^2+\frac1{r_2^2}\|x_2\|^2+b\|y\|^2-\frac2{r_1r_2}\langle x_1,x_2\rangle-2b\langle y,z\rangle+b\|z\|^2\tag4.$$
My first solution attempts was setting $$\sqrt bz=\frac{x_1}{r_1}+\frac{x_2}{r_2}.$$ With this $(2)$ becomes $$\frac2{r_1^2}\|x_1\|^2+\frac2{r_2^2}\|x_2\|^2+b\|y\|^2-2\sqrt b\left\langle y,\frac{x_1}{r_1}+\frac{x_2}{r_2}\right\rangle\tag5.$$
Let $\,\lambda = \dfrac{1}{\sigma_1^2}\,$, $\,\mu = \dfrac{1}{\sigma_2^2}$ and assume WLOG that $\,\lambda + \mu = 1\,$. Then:
$$ \begin{align} \lambda \|y - x_1\|^2 + \mu \|y - x_2 \|^2 &= \| y \|^2 - 2 \left\langle y, \lambda x_1 + \mu x_2\right\rangle + \lambda \|x_1\|^2+ \mu \|x_2\|^2 \\ &= \|y - (\lambda x_1 + \mu x_2)\|^2 - \|\lambda x_1 + \mu x_2\|^2 + \lambda \|x_1\|^2+ \mu \|x_2\|^2 \end{align} $$
For $(1) \equiv (2)$ to hold for all $y$ this requires $b=1, z = \lambda x_1 + \mu x_2$. The remaining terms:
$$ \begin{align} - \|\lambda x_1 + \mu x_2\|^2 + \lambda \|x_1\|^2+ \mu \|x_2\|^2 &= -\left(\lambda^2 \|x_1\|^2 + \mu^2\| x_2 \|^2 + 2 \lambda\mu \left\langle x_1, x_2 \right\rangle\right) \\ &\;\;\;\; + \lambda \|x_1\|^2+ \mu \|x_2\|^2 \\ &= \lambda \underbrace{(1-\lambda)}_{=\, \mu} \| x_1\|^2 + \mu\underbrace{(1 - \mu)}_{=\,\lambda} \| x_2\|^2 - 2 \lambda\mu \left\langle x_1, x_2 \right\rangle \\ &= \lambda\mu \left( \|x_1\|^2 + \|x_2\|^2 - 2 \left\langle x_1, x_2 \right\rangle \right) \\ &= \lambda\mu \|x_1 - x_2 \|^2 \end{align} $$
This implies $\,r_1 = r_2 = \dfrac{1}{\sqrt{\lambda\mu}}\,$.
May be worth noting that $\lambda \|y - x_1\|^2 + \mu \|y - x_2 \|^2 = \|y - (\lambda x_1 + \mu x_2)\|^2 + \lambda\mu \|x_1 - x_2 \|^2$ is equivalent to Stewart's Theorem for the triangle formed by $y, x_1, x_2$ and the point $\lambda x_1 + \mu x_2$ on the side $x_1x_2$.