I remember learning about this in design class. Each bolt has two components to it's shear force. One is equal to a fraction of the applied load directly, and the other is the shear force needed to create an equipollent moment at the center of the bolt pattern.
Specifically, with $n$ bolts, and $\mathbf{r}$ the location of the load from the bolt pattern center, the equipollent moment is
$$ M =\| \mathbf{r} \times \mathbf{F} \| = r_x F_y - r_y F_x$$
If the position of the i-th bolt is $\mathbf{b}_i = R_i \mathbf{e}_i$ where $R_i$ is the radial distance and $\mathbf{e}_i$ the radial direction vector.
Consider also the perpendicular direction $\mathbf{n}_i$. This points tangentially from each bolt.
The total shear force for the i-th bolt is
$$\mathbf{S}_i = \frac{1}{n} \left( \mathbf{F} + \frac{M}{R_i} \mathbf{n}_i \right) $$
You can show that this satisfies the equilibrium equations
$$\begin{cases} \sum \limits_{i=1}^n \mathbf{S}_i = \mathbf{F} \\ \sum \limits_{i=1}^n \| \mathbf{b}_i \times \mathbf{S}_i \| = M \end{cases} $$
This assumes that the connecting parts are compliant enough to give an equal load distribution to the bolts and that the bolt holes are a loose fit and don't push against the bolts when not loaded.