  Let $\mathcal{H}_q$ denote the Hermitian curve in $\mathbb{P}^2$ over
$\mathbb{F}_{q^2}$ and $\mathcal{C}_d$ be an irreducible plane projective curve
in $\mathbb{P}^2$ also defined over $\mathbb{F}_{q^2}$ of degree $d$. Can
$\mathcal{H}_q$ and $\mathcal{C}_d$ intersect in exactly $d(q+1)$ distinct
$\mathbb{F}_{q^2}$-rational points? B\'ezout's theorem immediately implies that
$\mathcal{H}_q$ and $\mathcal{C}_d$ intersect in at most $d(q+1)$ points, but
equality is not guaranteed over $\mathbb{F}_{q^2}$. In this paper we prove that
for many $d \le q^2-q+1$, the answer to this question is affirmative. The case
$d=1$ is trivial: it is well known that any secant line of $\mathcal{H}_q$
defined over $\mathbb{F}_{q^2}$ intersects $\mathcal{H}_q$ in $q+1$ rational
points. Moreover, all possible intersections of conics and $\mathcal{H}_q$ were
classified by Donati et al. in 2009 and their results imply that the answer to
the question above is affirmative for $d=2$ and $q \ge 4$, as well. However, an
exhaustive computer search quickly reveals that for $(q,d) \in
\{(2,2),(3,2),(2,3)\}$, the answer is instead negative. We show that for $q \le
d \le q^2-q+1$, $d=\lfloor(q+1)/2\rfloor$ and $d=3$, $q \geq 3$ the answer is
again affirmative. Various partial results for the case $d$ small compared to
$q$ are also provided.
