2. Given $a, b \in \mathbf{R}$, the circle $C_{1}: x^{2}+y^{2}-2 x+4 y-b^{2}+5=0$ intersects with $C_{2}: x^{2}+y^{2}-2(a-6) x-2 a y$ $+2 a^{2}-12 a+27=0$ at two distinct points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}+\frac{x_{1}-x_{2}}{y_{1}-y_{2}}$ $=0$, then $a=$ $\qquad$
2. $a=4$ Detailed explanation: Let the midpoint of $A B$ be $M$, and the origin of coordinates be $O$. Then, from $\frac{y_{1}+y_{2}}{x_{1}+x_{2}}+\frac{x_{1}-x_{2}}{y_{1}-y_{2}}=0$ we get $M O \perp A B$, so $O, M, C_{1}, C_{2}$ are collinear. From $k_{\alpha_{1}}=k_{\alpha_{2}}$ we get $a=4$.