Consider a hyperbola $$H$$ having centre at the origin and foci on the $$x$$-axis. Let $$C_1$$ be the circle touching the hyperbola $$H$$ and having the centre at the origin. Let $$C_2$$ be the circle touching the hyperbola $$H$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36\pi$$ and $$4\pi$$, respectively, then the length (in units) of latus rectum of $$H$$ is

A hyperbola $$H$$ centered at the origin with foci on the x-axis is given, and a circle $$C_1$$ centered at the origin touches $$H$$ and has area $$36\pi$$ while another circle $$C_2$$ centered at one focus touches $$H$$ at its vertex and has area $$4\pi$$.

From the areas we get $$r_1^2 = 36 \implies r_1 = 6$$ and $$r_2^2 = 4 \implies r_2 = 2$$.

The circle $$C_1$$ is tangent to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ at its closest point to the origin, namely the vertex $$(a,0)$$, so $$r_1 = a = 6$$.

Since the circle $$C_2$$ is centered at the focus $$(ae,0)$$ and touches the hyperbola at the vertex $$(a,0)$$, its radius is the distance from the focus to the vertex: $$r_2 = ae - a = a(e - 1) = 6(e - 1) = 2 \implies e - 1 = \frac{1}{3} \implies e = \frac{4}{3}$$.

Then $$b^2 = a^2(e^2 - 1) = 36\left(\frac{16}{9} - 1\right) = 36 \times \frac{7}{9} = 28$$.

Finally, the length of the latus rectum is $$\ell = \frac{2b^2}{a} = \frac{2 \times 28}{6} = \frac{56}{6} = \frac{28}{3}$$, so the correct answer is Option B: $$\frac{28}{3}$$.