8. Given the incenter $I(-1,7)$ of a right-angled triangle $\triangle O A B$ with all three vertices as integer points, and the origin $O$ as the right-angle vertex. The number of such right-angled triangles $\triangle O A B$ is $\qquad$

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8.2.

As shown in Figure 4.
Let $\angle x O A=\alpha, \angle x O I=\beta$.
Then $\alpha=\beta-\frac{\pi}{4}$.
Also, $\tan \beta=-7$, so
$$
\begin{array}{l}
\tan \alpha=\tan \left(\beta-\frac{\pi}{4}\right) \\
=\frac{\tan \beta-1}{1+\tan \beta}=\frac{4}{3}, \\
\tan \angle x O B=\tan \left(\frac{\pi}{2}+\alpha\right) \\
=-\cot \alpha=-\frac{1}{\tan \alpha}=-\frac{3}{4} .
\end{array}
$$

Then the point $A(3 t, 4 t), B(-4 k, 3 k)\left(k \backslash t \in \mathbf{Z}_{+}\right)$.
Thus, $O A=5 t, O B=5 k$, and the inradius of the right triangle $\triangle O A B$ is
$$
\begin{array}{l}
r=O I \sin 45^{\circ}=\frac{\sqrt{2}}{2} O I=5 . \\
\text { Also } r=\frac{1}{2}(O A+O B-A B) \\
\Rightarrow \frac{1}{2}\left[5 t+5 k-\sqrt{(5 t)^{2}+(5 k)^{2}}\right]=5 \\
\Rightarrow t+k-2=\sqrt{t^{2}+k^{2}} \\
\Rightarrow(t-2)(k-2)=2 .
\end{array}
$$

Since $t$ and $k$ are both positive integers, we have $(t, k)=(3,4)$ or $(4,3)$.
Therefore, there are two right triangles $\triangle O A B$ that satisfy the conditions.