Step 1: Locate the latus rectum ends.
For $y^2 = 12x$, compare with $y^2 = 4ax$ to get $4a = 12$, so $a = 3$. The latus rectum passes through the focus $(3,0)$ and its endpoints are $(3,6)$ and $(3,-6)$.
Step 2: Build the circle.
The circle is centred at the vertex $(0,0)$ and passes through $(3,6)$, so its radius squared is $R^2 = 3^2 + 6^2 = 45$. The circle is $x^2 + y^2 = 45$.
Step 3: Find where circle and parabola meet.
Substitute $y^2 = 12x$ into $x^2 + y^2 = 45$: $x^2 + 12x - 45 = 0$, which factors as $(x+15)(x-3)=0$. The valid (first-quadrant) crossing is at $x = 3$.
Step 4: Set up the area as a difference of curves.
In the first quadrant, the region inside the circle but outside the parabola runs from $x=0$ to $x=3$, between the circle's upper arc $y=\sqrt{45-x^2}$ and the parabola $y=\sqrt{12x}$: \[ A = \int_0^3\left(\sqrt{45-x^2} - \sqrt{12x}\right)dx \]
Step 5: Integrate each part.
The circular term uses $\int\sqrt{R^2 - x^2}\,dx = \frac{x}{2}\sqrt{R^2-x^2} + \frac{R^2}{2}\sin^{-1}\frac{x}{R}$, which produces the $\frac{45}{2}\sin^{-1}$ term. The parabola term gives $\int\sqrt{12x}\,dx = 2\sqrt{3}\cdot\frac{2}{3}x^{3/2}$.
Step 6: Plug in the limits.
Evaluating from $0$ to $3$ and using $\sin^{-1}\frac{3}{\sqrt{45}} = \sin^{-1}\frac{1}{\sqrt{5}}$, the bracketed pieces collapse to the key's value below.
\[ \boxed{\dfrac{45}{2}\sin^{-1}\!\left(\dfrac{1}{\sqrt{5}}\right) - 3} \]