4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 1, \\ 16 x^{4}-8 x^{2} y^{2}+y^{4}-40 x^{2}-10 y^{2}+25=0 .\end{array}\right.$
Answer. $\left(-\frac{2}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right),\left(-\frac{2}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{2}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{2}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right)$.

Solution. Transform the equation of the system:

$$
\begin{gathered}
16 x^{4}-8 x^{2} y^{2}+y^{4}-40 x^{2}-10 y^{2}+25=0 \Leftrightarrow 16 x^{4}+8 x^{2} y^{2}+y^{4}-40 x^{2}-10 y^{2}+25=16 x^{2} y^{2} \Leftrightarrow \\
\Leftrightarrow\left(4 x^{2}+y^{2}\right)^{2}-10\left(4 x^{2}+y^{2}\right)+25=16 x^{2} y^{2} \Leftrightarrow\left(4 x^{2}+y^{2}-5\right)^{2}=(4 x y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } 
{ 4 x ^ { 2 } + y ^ { 2 } - 5 = 4 x y , } \\
{ 4 x ^ { 2 } + y ^ { 2 } - 5 = - 4 x y }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ ( 2 x - y ) ^ { 2 } = 5 , } \\
{ ( 2 x + y ) ^ { 2 } = 5 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
2 x-y= \pm \sqrt{5}, \\
2 x+y= \pm \sqrt{5} .
\end{array}\right.\right.\right.
\end{gathered}
$$

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

If $y=2 x+\sqrt{5}$, then $x^{2}+(2 x+\sqrt{5})^{2} \leq 1,5 x^{2}+4 x \sqrt{5}+4 \leq 0,(x \sqrt{5}+2)^{2} \leq 0, x=-\frac{2}{\sqrt{5}}$. Then $y=\frac{1}{\sqrt{5}}$.

The other cases are considered similarly. In the end, we get 4 solutions: $\left(-\frac{2}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right),\left(-\frac{2}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right)$, $\left(\frac{2}{\sqrt{5}} ;-\frac{1}{\sqrt{5}}\right),\left(\frac{2}{\sqrt{5}} ; \frac{1}{\sqrt{5}}\right)$.