Vector Identity that appears in the Poisson Kernel of the $B(0,1)$

Question

Given that $y\in \partial B(0,1)$ how do i get that $|x|^2|y-\frac{x}{|x|^2}|=|y-x|$? It may be a silly question but I don't see why?

Thanks!

Answer 1

$$ |y-x|^2=|y|^2-2\,y\cdot x+|x|^2=1-2\,y\cdot x+|x|^2. $$ $$ |x|^2\,\Bigl|\,y-\frac{x}{|x|^2}\Bigr|^2=|x|^2\Bigl(|y|^2-\frac{2\,y\cdot x}{|x|^2}+\frac{1}{|x|^2}\Bigr)=|x|^2-2\,y\cdot x+1. $$