2. In the Cartesian coordinate system, the circle $\Omega: x^{2}+y^{2}+d x+e y+f=0$ (where $d, e, f$ are real numbers) has a diameter $A B$, where $A(20,22), B(10,30)$, then the value of $f$ is $\qquad$ .
Given the center $C(15,26) \Rightarrow r^{2}=5^{2}+4^{2}=41$, then the circle $\Omega:(x-15)^{2}+(y-26)^{2}=41$ $\Rightarrow f=15^{2}+26^{2}-41=225+676-41=860$.