双曲线 $C_1$ 的方程为 $\frac{(x-2)^2}{3^2} - \frac{y^2}{2^2} = 1$。

1. Find the axial intercepts of the hyperbola.

To find the $x$-intercepts, set $y = 0$:
$\frac{(x-2)^2}{3^2} - 0 = 1$
$(x-2)^2 = 9$
$x - 2 = \pm 3$
$x = 5 \quad \text{or} \quad x = -1$
The $x$-intercepts are $(5, 0)$ and $(-1, 0)$.

To find the $y$-intercepts, set $x = 0$:
$\frac{(-2)^2}{9} - \frac{y^2}{4} = 1$
$\frac{4}{9} - \frac{y^2}{4} = 1$
$-\frac{y^2}{4} = \frac{5}{9}$
Since $y^2 \ge 0$, there are no real solutions. Hence, there are no $y$-intercepts.

2. Determine the equations of the asymptotes.

The hyperbola is in the standard form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ with center $(h, k) = (2, 0)$, $a = 3$, and $b = 2$.
The equations of the asymptotes are given by $y - k = \pm \frac{b}{a}(x - h)$.
Substituting the values:
$y - 0 = \pm \frac{2}{3}(x - 2)$
So the asymptotes are:
$y = \frac{2}{3}(x - 2) \quad \text{and} \quad y = -\frac{2}{3}(x - 2)$

3. Sketch the graph incorporating all the key features.

1. Draw a set of coordinate axes and mark the center at $(2, 0)$.
2. Draw the dashed asymptotic lines $y = \frac{2}{3}(x - 2)$ and $y = -\frac{2}{3}(x - 2)$ passing through the center $(2,0)$.
3. Mark the vertices (which are also the $x$-intercepts) at $(-1, 0)$ and $(5, 0)$.
4. Sketch the two branches of the hyperbola opening horizontally from the vertices, approaching the asymptotes as $x \to \pm\infty$.

1. Use the geometric properties of the given hyperbola and circle to deduce the value of k for exactly two intersections.

The hyperbola $C_1$ and the circle $C_2$ both share the same centre at $(2, 0)$.
From the equation of $C_1$, the horizontal distance from the centre to its vertices is $a = 3$.
For the circle $C_2$ with radius $k$ to intersect the hyperbola exactly twice, it must be tangent to the two branches of the hyperbola at its vertices.
Therefore, the radius of the circle must exactly equal the distance from the centre to the vertices.
$k = 3$