6. Around an acute isosceles triangle $N P Q$ with base $N Q$, a circle $\Omega$ is circumscribed. The distance from the midpoint of the arc $P N$, not containing point $Q$, to the side $P N$ is 4, and the distance from the midpoint of the arc $Q N$, not containing point $P$, to the side $Q N$ is 0.4. Find the radius of the circle $\Omega$ and the area of triangle $N P Q$.
Answer: $R=5, S=\frac{192 \sqrt{6}}{25}$.

Solution. Let $C$ and $D$ be the midpoints of arcs $N P$ and $N Q$; $A$ and $H$ be the midpoints of segments $P N$ and $N Q$; $O$ be the center of circle $\Omega$, and $R$ be its radius. Then $O C \perp P N, O D \perp Q N ; O A=O C-A C=R-4, O H=O D-D H=R-\frac{2}{5}$.

From the Pythagorean theorem for triangles $O A N$ and $O H N$, we get that $A N^{2}=R^{2}-(R-4)^{2}=8 R-16$, $N D^{2}=R^{2}-\left(R-\frac{2}{5}\right)^{2}=\frac{4 R}{5}-\frac{4}{25}$. We find the sides of triangle $P N H: P H=P O+O H=2 R-\frac{2}{5}$, $P N=2 A N=2 \sqrt{8 R-16}, N H=\sqrt{\frac{4 R}{5}-\frac{4}{25}}$.

By the Pythagorean theorem for triangle $P N H$, we get $\left(2 R-\frac{2}{5}\right)^{2}+\left(\frac{4 R}{5}-\frac{4}{25}\right)=32 R-64$, from which $5 R^{2}-41 R+80=0, R=5$ or $R=\frac{16}{5}$. Only $R=5$ fits (in the second case, $A O=R-4<0$).

Thus, $P H=\frac{48}{5}, D H=\frac{4 \sqrt{6}}{5}, S_{N P Q}=P H \cdot N H=\frac{192 \sqrt{6}}{25}$.