AUT ABooklet

$$(x, y) \in \{(-2, 4), (-2, 6), (0, 10), (4, -2), (4, 12), (6, -2), (6, 12), (10, 0), (10, 10), (12, 4), (12, 6)\}.$$

An equivalent form of the given equation is
$$\begin{align*} x^2 + y^2 &= 10x + 10y \\ \iff x^2 - 10x + y^2 - 10y &= 0 \\ \iff (x - 5)^2 + (y - 5)^2 &= 50 \end{align*}$$
with $x + y \neq 0$. We therefore have to solve the equation $a^2 + b^2 = 50$ for integers $a$ and $b$.
As $\max(a^2, b^2) \leq 50$ we get $\max(|a|, |b|) \leq 7$. Furthermore we obtain $\min(a^2, b^2) \leq 25$ which yields $\min(|a|, |b|) \leq 5$. Analyzing the individual cases, we obtain $(a, b) \in \{(\pm1, \pm7), (\pm7, \pm1), (\pm5, \pm5)\}$ as the only possible solutions. If $(a, b) \in \{(\pm1, \pm7), (\pm7, \pm1)\}$, we get that $x - 5 = \pm1$ and $y - 5 = \pm7$ (or $x$ and $y$ swapped), which yields $x \in \{4, 6\}$ and $y \in \{-2, 12\}$ (or $y \in \{4, 6\}$ and $x \in \{-2, 12\}$). The case $(a, b) = (\pm5, \pm5)$ gives $x - 5 = \pm5$ and $y - 5 = \pm5$ which is equivalent to $x \in \{0, 10\}$ and $y \in \{0, 10\}$. The pair $(x, y) = (0, 0)$ is the only one violating $x + y \neq 0$. Therefore, we get the (eleven) different pairs listed in the answer.