11.3. Find the set formed by the solutions of the system of equations $\left\{\begin{array}{l}a x+y=2 a+3, \\ x-a y=a+4,\end{array}\right\}$ for all possible values of the parameter $a$. (For each value of $a$, a solution $(x, y)$ of the given system is found, and all these solutions together form the desired set of points on the coordinate plane.)
Answer. A circle with center $(3,1)$ and radius $\sqrt{5}$, with the point $(2,-1)$ removed.

Solution. Consider an arbitrary solution $(x, y)$ of the given system. If $x \neq 2$, express $a$ from the first equation in terms of $x, y: a=\frac{y-3}{2-x}$. Similarly, if $y \neq-1$, express $a$ from the second equation in terms of $x, y: a=\frac{x-4}{y+1}$. By equating these expressions, after transformations, we get $(x-3)^{2}+(y-1)^{2}=5$, which is the equation of a circle with center $(3,1)$ and radius $\sqrt{5}$. Therefore, all solutions of the system lie on this circle. On the other hand, for any point $(x, y)$ on this circle, where $x \neq 2$ and $y \neq-1$, we can rewrite the equation of the circle in the equivalent form $\frac{y-3}{2-x}=\frac{x-4}{y+1}$, and set $a$ equal to the value of this ratio. Then, rewrite the equalities $\frac{y-3}{2-x}=a, \frac{x-4}{y+1}=a$ equivalently to the original system and verify that the pair $x, y$ is its solution. Let's consider the remaining cases. If $x=2, y \neq-1$, then from the first equation $y=3$ for any $a$, and from the second $a=-\frac{1}{2}$ - for such $a$, the pair $(2,3)$ will be a solution of the system. If $x \neq 2, y=-1$, then from the second equation $x=4$ for any $a$, and from the first $a=2$ - for such $a$, the pair $(4,-1)$ will be a solution of the system. Both of these points lie on the circle. From these same arguments, it follows that the last remaining point $(2,-1)$ cannot be a solution of the system for any $a$.

Criterion. Equation of the circle: 3 points.

Shown that all points on the circle, except three, are solutions: 4 points.

Point $(2,-1)$ not removed - minus 2 points.

Extra points removed - minus 1 point for each.