Let the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with centre at origin and foci on the $x$-axis. Then its foci are at $(\pm c,0)$, where $c^2=a^2+b^2.$

We are given two circles.

1. Circle $C_1$

$C_1$ has centre at the origin and touches the hyperbola.

For the hyperbola, the nearest points to the origin are the vertices $(\pm a,0)$. Hence the circle centered at origin touching the hyperbola must have radius $r_1=a.$

Given area of $C_1$ is $36\pi$, so $\pi r_1^2=36\pi \implies r_1^2=36 \implies a=6.$

2. Circle $C_2$

$C_2$ has centre at one focus and touches the hyperbola at its vertex.

Take the focus $(c,0)$ and the vertex $(a,0)$. Since the circle touches the hyperbola at the vertex, its radius equals the distance from the focus to the vertex: $r_2=c-a.$

Given area of $C_2$ is $4\pi$, so $\pi r_2^2=4\pi \implies r_2=2.$ Thus, $c-a=2.$ Since $a=6$, $c=8.$

3. Find $b^2$

Using $c^2=a^2+b^2,$ we get $8^2=6^2+b^2$ $64=36+b^2$ $b^2=28.$

4. Length of latus rectum

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the length of latus rectum is $\frac{2b^2}{a}.$ So, $\text{latus rectum} = \frac{2\cdot 28}{6} = \frac{56}{6}=\frac{28}{3}.$

5. Option check

The value is $\frac{28}{3}$ which matches Option A.

Final Answer

$\boxed{\frac{28}{3}}$