Proving that a rectangle with specific vertices is in the opened unit circle In an exercise I'm asked to prove that:

Let $(a,b)$ be any point in $D=\{(x,y)\in \mathbb R ^2 : x^2 + y^2 < 1\}$. Put $r=\sqrt{a^2 + b^2}$. Let $R_{(a,b)}$ be the open rectangle with vertices at the points $\left(a\pm \frac{1-r}{8}, b\pm \frac{1-r}{8}\right)$. Prove that for every point $(a,b)$, $R_{(a,b)}\subset D$.

My first attempt was to calculate the distance from the center to each Vertex and show that it is less that $1$, thus being an element of the set $D$, but I failed. How can I prove this?

Given that the rectangle is effectively a square of side $l=(1-r)/4$, let us consider the smallest circle containing the square, whose radius will be $\rho=l/\sqrt{2}=(1-r)/(4\sqrt{2})$, and whose center $C$ is the center of the square.
The small circle, and so the square, is contained in the unit circle if and only if
$$
\overline{OC}+\rho\leq1\quad\implies\quad\rho\leq1-r
$$
i.e.
$$
\frac{1}{4\sqrt{2}}\leq1
$$
and this is true.
We can further observe that if we start with $(1-r)/n$ instead of $(1-r)/8$, the condition on $n$ is
$$
\frac{2}{n\sqrt{2}}\leq1\quad\implies\quad n\geq\sqrt{2},
$$ so that the number $8$ is too large.
