4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 2, \\ 81 x^{4}-18 x^{2} y^{2}+y^{4}-360 x^{2}-40 y^{2}+400=0 .\end{array}\right.$
Answer. $\left(-\frac{3}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right),\left(-\frac{3}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{3}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{3}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right)$.

Solution. Transform the equation of the system:

$$
\begin{gathered}
81 x^{4}-18 x^{2} y^{2}+y^{4}-360 x^{2}-40 y^{2}+400=0 \Leftrightarrow 81 x^{4}+18 x^{2} y^{2}+y^{4}-360 x^{2}-40 y^{2}+400=36 x^{2} y^{2} \Leftrightarrow \\
\Leftrightarrow\left(9 x^{2}+y^{2}\right)^{2}-40\left(9 x^{2}+y^{2}\right)+400=36 x^{2} y^{2} \Leftrightarrow\left(9 x^{2}+y^{2}-20\right)^{2}=(6 x y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } 
{ 9 x ^ { 2 } + y ^ { 2 } - 2 0 = 6 x y , } \\
{ 9 x ^ { 2 } + y ^ { 2 } - 2 0 = - 6 x y }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ ( 3 x - y ) ^ { 2 } = 2 0 , } \\
{ ( 3 x + y ) ^ { 2 } = 2 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
3 x-y= \pm 2 \sqrt{5}, \\
3 x+y= \pm 2 \sqrt{5} .
\end{array}\right.\right.\right.
\end{gathered}
$$

In each of the four cases, express $y$ and substitute into the inequality.

If $y=3 x+2 \sqrt{5}$, then $x^{2}+(3 x+2 \sqrt{5})^{2} \leq 2,10 x^{2}+12 x \sqrt{5}+18 \leq 0,(x \sqrt{10}+3 \sqrt{2})^{2} \leq 0, x=-\frac{3}{\sqrt{5}}$. Then $y=\frac{1}{\sqrt{5}}$.

The other cases are considered similarly. In the end, we get 4 solutions: $\left(-\frac{3}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right),\left(-\frac{3}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right)$, $\left(\frac{3}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{3}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right)$.