## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-293.jpg?height=250&width=320&top_left_y=479&top_left_x=172)
Let $O$ be the orthocenter of $\triangle ABC$. According to Theorem 8, the circle passing through points $B, N, L$ will also pass through $O$. Moreover, by Theorem 6, $BO$ is its diameter $\left(\angle BNO = 90^\circ\right)$. The desired radius is $\frac{BO}{2}$.

$\triangle ANC \sim \triangle BNO \Rightarrow \frac{BO}{AC} = \frac{NO}{AN} = \tan \angle NAO = \tan(90^\circ - \angle B) = \cot \alpha \Rightarrow \frac{BO}{2} = \frac{a}{2} \cot \alpha$.

Answer: $\frac{1}{2} a \cdot \cot \alpha$.
12.410 In an acute-angled triangle $A B C$, altitudes $A L$ and $C N$ are drawn. Find the radius of the circle passing through points $B, L$, and $N$, if $A C=a$ and $\angle A B C=\alpha$.