4. Solve the system $\left\{\begin{array}{l}x^{2}+y^{2} \leq 1, \\ x^{4}-18 x^{2} y^{2}+81 y^{4}-20 x^{2}-180 y^{2}+100=0 .\end{array}\right.$
Answer. $\left(-\frac{1}{\sqrt{10}} ; \frac{3}{\sqrt{10}}\right),\left(-\frac{1}{\sqrt{10}} ;-\frac{3}{\sqrt{10}}\right),\left(\frac{1}{\sqrt{10}} ; \frac{3}{\sqrt{10}}\right),\left(\frac{1}{\sqrt{10}} ;-\frac{3}{\sqrt{10}}\right)$.

Solution. Transform the equation of the system:

$$
\begin{gathered}
x^{4}-18 x^{2} y^{2}+81 y^{4}-20 x^{2}-180 y^{2}+100=0 \Leftrightarrow x^{4}+18 x^{2} y^{2}+81 y^{4}-20 x^{2}-180 y^{2}+100=36 x^{2} y^{2} \Leftrightarrow \\
\Leftrightarrow\left(x^{2}+9 y^{2}\right)^{2}-20\left(x^{2}+9 y^{2}\right)+100=36 x^{2} y^{2} \Leftrightarrow\left(x^{2}+9 y^{2}-10\right)^{2}=(6 x y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } 
{ x ^ { 2 } + 9 y ^ { 2 } - 1 0 = 6 x y , } \\
{ x ^ { 2 } + 9 y ^ { 2 } - 1 0 = - 6 x y }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ ( x - 3 y ) ^ { 2 } = 1 0 , } \\
{ ( x + 3 y ) ^ { 2 } = 1 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x-3 y= \pm \sqrt{10}, \\
x+3 y= \pm \sqrt{10} .
\end{array}\right.\right.\right.
\end{gathered}
$$

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

If $x=3 y+\sqrt{10}$, then $y^{2}+(3 y+\sqrt{10})^{2} \leq 1,10 y^{2}+6 y \sqrt{10}+9 \leq 0,(y \sqrt{10}+3)^{2} \leq 0, y=-\frac{3}{\sqrt{10}}$. Then $x=\frac{1}{\sqrt{10}}$.

The other cases are considered similarly. In the end, we get 4 solutions: $\left(-\frac{1}{\sqrt{10}} ; \frac{3}{\sqrt{10}}\right)$, $\left(-\frac{1}{\sqrt{10}} ;-\frac{3}{\sqrt{10}}\right),\left(\frac{1}{\sqrt{10}} ; \frac{3}{\sqrt{10}}\right),\left(\frac{1}{\sqrt{10}} ;-\frac{3}{\sqrt{10}}\right)$.