(2) Draw two tangent lines to the circle $x^{2}+y^{2}=1$ through the point $(1,2)$. The area of the quadrilateral formed by these two tangent lines with the $x$-axis and $y$-axis is $\qquad$ .
(2) $\frac{13}{8}$ Hint: Clearly one of the tangent points is $(1,0)$, and the corresponding tangent segment length is 2, forming a trapezoid. Let the other tangent point be $\left(x_{0}, y_{0}\right)$, then we have $\left(x_{0}-1\right)^{2}+$ $\left(y_{0}-2\right)^{2}=4$ and $x_{0}^{2}+y_{0}^{2}=1$. Solving these equations yields $\left(x_{0}, y_{0}\right)=\left(-\frac{3}{5}, \frac{4}{5}\right)$, and the corresponding tangent line equation is $y=\frac{3}{4}(x-1)+2$. The intersection point of this tangent line with the $y$-axis is $\left(0, \frac{5}{4}\right)$, so the area of the trapezoid is $S=\frac{1}{2}\left(2+\frac{5}{4}\right)=\frac{13}{8}$.