Since $0 \leq (1 - x^2) \leq 1$, $0 \leq (1 - y^2) \leq 1$, we get that
$$$$
\sqrt{1-x^2} + \sqrt{1-y^2} \geq (1-x^2) + (1-y^2), \text{ and so the equality holds if }
\begin{cases}
\sqrt{1-x^2} = 1-x^2 \\
\sqrt{1-y^2} = 1-y^2
\end{cases} \Rightarrow
\begin{cases}
x^2 \in \{0;1\} \\
y^2 \in \{0;1\}
\end{cases}.
\text{ Thus we get the points from the answer.}