The larger of the two areas into which the circle  $x^2+y^2=64a^2$  is divided by the parabola  $y^2=12axis$

$x^2+y^2=64a^2$  is a circle with centre (0, 0) and radius 8a and  $y^2=12ax$  is a parabola whose vertex is at (0, 0) and latus rectum 12a. Both the curves are symmetrical about 
x-axis. Solving the two equations, the co-ordinates of the common point P are  $(4a,4a\sqrt{3})$.

Now the area of the larger portion of the circle (i.e. the shaded area) = the area PRSTQOP = the area of the semi-circle RST+2 area OPR
$$\begin{gathered} =\frac{1}{2}.\pi \left(8a^2\right)+[areaOPM+areaMPR] \\ =\frac{1}{2}.\pi {\left(8a\right)}^2+2[{\int }_0^{4a\sqrt{3}}xdy,for{y}^2=12ax+2[{\int }_{4a\sqrt{3}}^{3a}xdy,for{x}^2+y^2=64a^2] \\ =32\pi a^2+2{\int }_0^{4a\sqrt{3}}\frac{y^2}{12a}dy+2{\int }_{4a\sqrt{3}}^{8a}\sqrt{(4a^2-y^2}dy \\ =32\pi a^2+\frac{1}{6a}{\left[\frac{y^3}{3}\right]}_0^{4a\sqrt{3}}+2{[\frac{1}{2}y\sqrt{(64a^2-y^2)}+\frac{64a^2}{2}{\sin }^{-1}\frac{y}{8a}]}_{4a\sqrt{3}}^{8a} \\ =32\pi a^2+\frac{1}{6a}\left[\frac{64\times 3\sqrt{3}{a}^3}{3}\right]+2\left[\{0-8a^2\sqrt{3}\}+32a^2\{{\sin }^{-1}-{\sin }^{-1}(\sqrt{3/2}\left)\right\}\right] \\ =32\pi a^2+\frac{32\sqrt{3}{a}^2}{3}-16a^2\sqrt{3}+\frac{32}{3}{a}^2\pi \\ =\frac{128}{3}{a}^2\pi -\frac{16}{3}{a}^2\sqrt{3}=\frac{16}{3}{a}^2(8\pi -\sqrt{3}) \end{gathered}$$