Where do I look at this from? (1)

the title must have given u a hint that transformation of axes may help

let's set $\displaystyle \large X = xcos\theta+ysin\theta-1$

$\displaystyle \large Y = ycos\theta-xsin\theta-1$

this physically,is moving the origin of co-ordinate axes to $(1,1)$ and then rotating the axes by an angle $\theta$

$\displaystyle \large S_1$ in the new referrence frame is $\displaystyle \large X^{2}+Y^{2}-1=0$ and the point P is $(-2,0)$

from here it is plain geometry

let O be centre of $S_{1}$ and E be the point of tangency of AP with $S_{1}$ and F be the point of tangency of BP with $S_{1}$

angle APB is $\pi /3$

if D be the point of meeting of circles, angle APD is $\pi /6$

we find area of sector APD and then subtract half of the area of $S_1 = \frac{\pi}{2}$ subtract the area of triangle $OPE = \frac{\sqrt{3}}{2}$ and add half the area of sector $OEF = \frac{\pi}{6}$

this gives us $\displaystyle \large \frac{5\pi}{12}-\frac{\sqrt{3}}{2}$

I did it by a bit lengthier method. I did it by calculating areas. First I found out the distance between centres of two circles. Then, I found the radius of given Circle, S1. Now that I got these two, I found the angle between tangents. So I calculated the area enclosed between the tangents and circle, S1. Now, I calculated the area of the bigger sector in which the smaller circle, S1 is present. I equated this area to sum of area between tangents and smaller circle, S1 + area of smaller circle S1+ two leftover areas(which are equal because of symmetry). I solved this equation and got the area. But still, this method is a bit lengthy. Rohith M.Athreya's solution is unique and simple.