= Solution
Set $w=(x-y)/2$. The sum condition then gives $x=a+w$ and $y=a-w$, and their <dot product> is
$$
x\cdot y=(a+w)\cdot(a-w)=|a|^2-|w|^2.
$$
Therefore $|w|^2=|a|^2-c$. Since $c<|a|^2$, this is a strictly positive squared radius. The complete solution is
$$
\boxed{x=a+Rn,\qquad y=a-Rn,\qquad R=\sqrt{|a|^2-c},\qquad |n|=1.}
$$
Every choice of <unit vector> $n$ satisfies both conditions, and every solution has this form. The endpoints lie on the <sphere> with center $a$ and radius $R$, and their midpoint is $a$. They are therefore opposite ends of one diameter of that <sphere>.
Back to article page
