2. Draw two tangent lines to the circle $x^{2}+y^{2}=1$ through the point $(1,2)$. Then the area of the quadrilateral formed by these two tangent lines with the $x$-axis and $y$-axis is $\qquad$
2. $\frac{13}{8}$.

Obviously, one of the tangent points is $(1,0)$, and the corresponding tangent segment length is 2. Therefore, the quadrilateral formed is a trapezoid.
Let the other tangent point be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
\left(x_{0}-1\right)^{2}+\left(y_{0}-2\right)^{2}=4, \\
x_{0}^{2}+y_{0}^{2}=1
\end{array}\right. \\
\Rightarrow\left(x_{0}, y_{0}\right)=\left(-\frac{3}{5}, \frac{4}{5}\right) .
\end{array}
$$

Thus, the corresponding tangent line equation is $y=\frac{3}{4}(x-1)+2$, and the intersection point of this tangent line with the $y$-axis is $\left(0, \frac{5}{4}\right)$.
Therefore, the area $S$ of this trapezoid is $S=\frac{1}{2}\left(2+\frac{5}{4}\right)=\frac{13}{8}$.