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

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=26(y \sin a-x \cos a) \\
x^{2}+y^{2}=26(y \cos 2 a-x \sin 2 a)
\end{array}\right.
$$

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 24.
Answer: $a=\frac{\pi}{6} \pm \frac{2}{3} \operatorname{arctg} \frac{5}{12}+\frac{2 \pi n}{3}, \frac{\pi}{6}+\frac{2 \pi n}{3}$, where $n \in \mathbb{Z}$.

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

$$
\left\{\begin{array}{l}
(x+13 \cos a)^{2}+(y-13 \sin a)^{2}=169 \\
(x+13 \sin 2 a)^{2}+(y-13 \cos 2 a)^{2}=169
\end{array}\right.
$$

Each of these equations represents a circle with radius 13; the center of the first circle is the point $A(-13 \cos a ; 13 \sin a)$, and the center of the second circle is the point $B(-13 \sin 2 a ; 13 \cos 2 a)$.

If these equations represent the same circle, then there will be points on this circle that are 24 units apart. The circles coincide if they have the same center. We get

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

These equalities hold if either $\cos \frac{\pi+6 a}{4}=0$ or $\cos \frac{\pi+2 a}{4}=\cos \frac{\pi-2 a}{4}=0$. In the first case, we get $a=\frac{\pi}{6}+\frac{2 k \pi}{3}, k \in \mathbb{Z}$. 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 24 units apart, $P Q=24$, therefore $A B^{2}+576=4 \cdot 169, A B=10$. Thus, it is necessary that the distance between the centers of the circles $A$ and $B$ be 10. From this,

$$
\begin{aligned}
& \sqrt{(-13 \sin 2 a+13 \cos a)^{2}+(13 \cos 2 a-13 \sin a)^{2}}=10 \Leftrightarrow \\
& \quad \Leftrightarrow 338-338 \sin 2 a \cos a-338 \sin a \cos 2 a=100 \Leftrightarrow 338 \sin 3 a=238 \Leftrightarrow
\end{aligned}
$$

$$
\Leftrightarrow a=\frac{(-1)^{k}}{3} \arcsin \frac{119}{169}+\frac{k \pi}{3}, k \in \mathbb{Z}
$$