2. Given the circle $C:(x-1)^{2}+(y-2)^{2}=2$, point $P(2,-1)$, and the tangents $PA$ and $PB$ from $P$ to circle $C$, where $A$ and $B$ are the points of tangency, the equation of the line $AB$ is $\qquad$
2. $x-3 y+3=0$ Detailed Explanation: Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $\left(x_{1}-1\right)^{2}+\left(y_{1}-2\right)^{2}=2,\left(x_{2}-1\right)^{2}+\left(y_{2}-\right.$ $2)^{2}=2, \because \quad C A \perp A P, \therefore k_{C A} \cdot k_{A P}=-1, \therefore \frac{y_{1}-2}{x_{1}-1} \cdot \frac{y_{1}+1}{x_{1}-2}=-1, \therefore\left(y_{1}-2\right)\left(y_{1}+1\right)=-\left(x_{1}-1\right)$ $\left(x_{1}-2\right), \therefore\left(y_{1}-2\right)\left(y_{1}-2+3\right)=-\left(x_{1}-1\right)\left(x_{1}-1-1\right)$, i.e., $\left(y_{1}-2\right)^{2}+3\left(y_{1}-2\right)=-\left(x_{1}-1\right)^{2}+$ $\left(x_{1}-1\right), \therefore\left(x_{1}-1\right)^{2}+\left(y_{1}-2\right)^{2}+3\left(y_{1}-2\right)-\left(x_{1}-1\right)=0, \because\left(x_{1}-1\right)^{2}+\left(y_{1}-2\right)^{2}=2, \therefore x_{1}-3 y_{1}+$ $3=0$, Similarly, we get $x_{2}-3 y_{2}+3=0, \because A 、 B$ two points' coordinates both satisfy the equation $x-3 y+3=0, \therefore$ the equation of line $A B$ is $x-3 y+3=0$.