6. Find all values of the parameter $a$, for each of which the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=10(x \cos 4 a+y \sin 4 a) \\
x^{2}+y^{2}=10(x \sin a+y \cos a)
\end{array}\right.
$$

has two solutions $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ such that the distance between the points $P\left(x_{1} ; y_{1}\right)$ and $Q\left(x_{2} ; y_{2}\right)$ is 6.
Answer: $a=\frac{\pi}{10} \pm \frac{2}{5} \operatorname{arctg} \frac{4}{3}+\frac{2 \pi n}{5}, \frac{\pi}{10}+\frac{2 \pi n}{5}$, where $n \in \mathbb{Z}$.

Solution. By completing the squares, we rewrite the system as

$$
\left\{\begin{array}{l}
(x-5 \cos 4 a)^{2}+(y-5 \sin 4 a)^{2}=25 \\
(x-5 \sin a)^{2}+(y-5 \cos a)^{2}=25
\end{array}\right.
$$

Each of these equations represents a circle of radius 5; the center of the first circle is the point $A(5 \cos 4 a ; 5 \sin 4 a)$, and the center of the second circle is the point $B(5 \sin a ; 5 \cos a)$.

If these equations represent the same circle, then there will be points on this circle that are 6 units apart. The circles coincide when they have the same center. We get

$$
\left\{\begin{array} { l } 
{ \operatorname { c o s } 4 a = \operatorname { s i n } a , } \\
{ \operatorname { s i n } 4 a = \operatorname { c o s } a }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \operatorname { c o s } 4 a + \operatorname { c o s } ( \frac { \pi } { 2 } + a ) = 0 , } \\
{ \operatorname { c o s } a + \operatorname { c o s } ( \frac { \pi } { 2 } + 4 a ) = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 \cos \frac{\pi+10 a}{4} \cos \frac{\pi-6 a}{4}=0 \\
2 \cos \frac{\pi+10 a}{4} \cos \frac{\pi+6 a}{4}=0
\end{array}\right.\right.\right.
$$

These equalities hold if either $\cos \frac{\pi+10 a}{4}=0$ or $\cos \frac{\pi-6 a}{4}=\cos \frac{\pi+6 a}{4}=0$. In the first case, we get $a=\frac{\pi}{10}+\frac{2 k \pi}{5}, k \in \mathbb{Z}$. The second case clearly does not provide solutions.

Now suppose the circles are different and intersect at points $P$ and $Q$. Then the quadrilateral $A P B Q$ is a rhombus. It is known that in any parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of all four sides, so $A B^{2}+P Q^{2}=4 A P^{2}$. Since we want the points $P$ and $Q$ to be 6 units apart, $P Q=6$, therefore $A B^{2}+36=4 \cdot 25, A B=8$. Thus, it is necessary that the distance between the centers of the circles $A$ and $B$ be 8. From this,

$$
\begin{aligned}
& \sqrt{(5 \sin a-5 \cos 4 a)^{2}+(5 \cos a-5 \sin 4 a)^{2}}=8 \Leftrightarrow \\
& \Leftrightarrow 50-50 \sin a \cos 4 a-50 \sin 4 a \cos a=64 \Leftrightarrow 50 \sin 5 a=-14 \Leftrightarrow \\
& \Leftrightarrow a=\frac{(-1)^{k+1}}{5} \arcsin \frac{7}{25}+\frac{k \pi}{5}, k \in \mathbb{Z}
\end{aligned}
$$