Show open rectangles with vertices $\langle a\pm\frac{1-r}{8},b\pm\frac{1-r}{8}\rangle $ are contained in the unit disk

Question

Let $D$ be the disk $\{\langle x,y \rangle: x^2+y^2<1\}$. I've been asked to show that if $\langle a,b \rangle \in D$, $r=\sqrt{a^2+b^2}$ and $R_{\langle a,b \rangle}$ is the open rectangle with vertices $\langle a\pm\frac{1-r}{8},b\pm\frac{1-r}{8}\rangle $ then $R_{\langle a,b \rangle}\subset D$. I have shown this via a trigonometric substitution and a quadratic inequity, however my solution was long and tedious and it doesn't really show where this rectangle came from. I do not understand what is unique about $\frac{1-r}{8}$ and i believe for larger rectangles, for example one with vertices $\langle a\pm\frac{1-r}{4},b\pm\frac{1-r}{4}\rangle $ The inequity would still work. If anyone could help find an intuitive solution which explains why $\frac{1- r}{8}$ is used that would be amazing.

Answer 1

Given $v=(a,b)\subset D $ we need to find $w_1=(k,k)$ and $w_2=(k,-k)$ such that $v\pm w_1=(a\pm k,b\pm k) \subset D$ and $v\pm w_2=(a\pm,b\mp k) \subset D$.

From simply geometrical properties it easy to verify that the maximum value for k which fullfils the conditions is $k=\frac{1-r}{\sqrt 2}$.

Indeed the worst case is atteined when $v$ and $w$ are positive multiple (i.e. aligned and in the same direction) thus

$$|v+w|=|v|+|w|<1\implies|w|<1-r\implies k \sqrt 2<1-r\implies k<\frac{1-r}{\sqrt 2} $$

Answer 2

This seems straightforward. $r$ is the distance from $(a,b)$ to the origin, so $1-r$ is the distance to the boundary. The largest disk centered at $(a,b)$ that will fit within the disk has radius $1-r,$ and the side of the inscribed square is $(1-r)/\sqrt(2).$ We need half the side, so any denominator $\ge 2\sqrt(2)$ will work.

I can't say why 8 is used. What was the context?