Solution
= Solution
With wall effects neglected, \b[$a_1=a$]. A second <hard disc> cannot have its centre within distance $2r$ of the first centre. Its <excluded volume> in this two-dimensional setting is an excluded area $\pi(2r)^2=4a_d$, giving
$$
\boxed{a_2=a-4a_d}.
$$
This is exclusion of the second centre, rather than simply subtraction of the first disc's own area.