3. [6 points] On the plane $O x y$, there is a point $A$, the coordinates $(x ; y)$ of which satisfy the equation $2 a^{2}+2 a x+x^{2}-2 x y+2 y^{2}=0$, and a circle with center at point $B$, given by the equation $a^{2} x^{2}+a^{2} y^{2}-$ $2 a^{3} x-6 a x-2 a^{2} y+a^{4}+9=0$. Find all values of the parameter $a$ for which points $A$ and $B$ lie on opposite sides of the line $x=4$ (points $A$ and $B$ do not lie on this line).
Answer: $(-\infty ;-2) \cup(0 ; 1) \cup(3 ; \infty)$.

Solution. The first equation can be transformed as follows:

$$
\begin{aligned}
&\left(x^{2}+2(a-y) x\right)+2 y^{2}+2 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}+2 \cdot x \cdot(a-y)+(a-y)^{2}\right)-(a-y)^{2}+2 y^{2}+2 a^{2}=0 \Leftrightarrow \\
& \Leftrightarrow(x+a-y)^{2}+y^{2}+2 a y+a^{2}=0 \Leftrightarrow(x+a-y)^{2}+(y+a)^{2}=0
\end{aligned}
$$

From this, we obtain that the coordinates of point $A\left(x_{A} ; y_{A}\right)$ satisfy the equations $x+a-y=0$ and $y+a=0$, i.e., $x_{A}=-2 a, y_{A}=-a$.

Since the second equation defines a circle, $a \neq 0$. Under this condition, the equation can be rewritten as $x^{2}+y^{2}-2 a x-\frac{6 x}{a}-2 y+a^{2}+\frac{9}{a^{2}}=0$, from which we further get $\left(x^{2}-2 x\left(a+\frac{3}{a}\right)+\left(a+\frac{3}{a}\right)^{2}\right)+\left(y^{2}-2 y+1\right)+a^{2}+$ $\frac{9}{a^{2}}=1+\left(\frac{3}{a}+a\right)^{2}$, and finally $\left(x-a-\frac{3}{a}\right)^{2}+(y-1)^{2}=7$. Therefore, the coordinates of point $B$ are: $x_{B}=a+\frac{3}{a}, y_{B}=1$.

For points $A$ and $B$ to lie on opposite sides of the line $x=4$, it is necessary and sufficient that their abscissas satisfy the inequality $\left(x_{A}-4\right)\left(x_{B}-4\right)0 \Leftrightarrow \frac{(a+2)(a-1)(a-3)}{a}>0$. Solving this inequality using the interval method, we get the answer: $a \in(-\infty ;-2) \cup(0 ; 1) \cup(3 ;+\infty)$.

Remark. To solve the problem, it is sufficient to find the abscissas of points $A$ and $B$.