8. In the Cartesian coordinate system $x O y$, $\Gamma_{1}$ is a unit circle with center at $(2,1)$, and $\Gamma_{2}$ is a unit circle with center at $(10,11)$. A line $l$ is drawn through the origin $O$ such that $l$ intersects $\Gamma_{1}$ and $\Gamma_{2}$ at two points each, dividing $\Gamma_{1}$ and $\Gamma_{2}$ into four arcs, with two of these arcs being equal in length. The sum of the slopes of all lines $l$ that satisfy this condition is $\qquad$
The solution is as follows:
As shown in the figure, there are three lines that satisfy the conditions: $l_{1}$ passes through point $B$, $l_{2}$ passes through the midpoint $C$ of line segment $A B$, and $l_{3} \parallel A B$.
Thus, $k_{l_{1}}=\frac{11}{10}, k_{l_{2}}=1, k_{l_{3}}=\frac{5}{4}$,
Therefore, the sum of the slopes of all lines $l$ that satisfy the conditions is
$$
\frac{11}{10}+1+\frac{5}{4}=\frac{67}{20} .
$$