<s>[INST] What is the area of shaded region which is lies between outer and inner circle. There is a outer circle with radius 2r and another inner circle with radius r whose center is the middle of big circle.As depicted in the following figure.

Foo graph Image
There is a sector of 120 degree in inner circle which leads to shaded part between outer and inner circle. What is the area of shaded circle in terms of r ?
i.e. What is the value of K ?
 [/INST] You can also easily solve the problem by rotating the figure by $90^o$ clockwise so that the dotted line coincides with the x-axis & center of outer circle at the origin. Thus, equation of the outer circle: $x^2+y^2=4r^2$ & the equation radius above x-axis: $y-0=\tan60^o(x-r)$ or $y=\sqrt{3}(x-r)$ solving the equation for intersection point of line & outer circle as follows $$x^2+(\sqrt{3}(x-r))^2=4r^2 => 4x^2-6rx-r^2$$ now, solving for x-coordinate, we get $$x=\frac{6r\pm\sqrt{36r^2+16r^2}}{8}=\frac{r(3\pm \sqrt{13})}{4}$$ Consider a right triangle above x-axis obtained by joining the intersection point (in I-quadrant) to the center of outer circle, we get semi-aperture angle $\alpha$ at the center of outer circle as follows $$\alpha=\cos^{-1}\left(\frac{\frac{r(3+\sqrt{13})}{4}}{2r}\right)=\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)=\sin^{-1}\left(\frac{\sqrt{42-6\sqrt{13}}}{8}\right)$$       
Hence, the shaded area is $$=\frac{1}{2}(2\alpha)(2r)^2-2\left(\frac{1}{2}(r)(2r)(\sin\alpha)\right)-\frac{1}{2}\left(\frac{2\pi}{3}\right)(r^2)=r^2\left(4\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)-\frac{\pi}{3}-\left(\frac{\sqrt{42-6\sqrt{13}}}{4}\right)\right)$$ Hence, we get $$K=4\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)-\frac{\pi}{3}-\frac{\sqrt{42-6\sqrt{13}}}{4}\approx0.222026268$$
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