<s>[INST] Area of a region delimited by chords and circular arcs. 
Let $AB$ be the diameter of circle $O$, where $AB = 2$. Circle $P$ is internally tangent to circle $O$ at point $B$, and $PB$ = $\frac{2}{3}$.
  Two different chords $AX$ and $AY$ are drawn tangent to circle
  P. Let $R$ be the region bounded by $AX, AY$ , and arc $XBY$ . What is the area of the region inside R but outside circle $P$?

I was able to figure out that the sides of the other two sides were $\sqrt 3$ and $1$ using similar triangles. Then the segments were $\frac{π}{6}$-$\frac{\sqrt 3}{4}$. I added the area of the triangle then multiplied it by two and subtracted the smaller circle's area from the result. My answer is $\frac{\sqrt 3}{2}$-$\frac{π}{9}$. But the correct answer is $\frac{4\sqrt 3}{9}$-$\frac{4π}{27}$. Is there anything wrong with my solution?
 [/INST] From the question AB = 2 and PB = 2/3.  Then AP = 4/3 and PX = 2/3.
Then $AX = \sqrt{AP^2 - PX^2} = \frac{2\sqrt3}{3}$
Sum of area 
$$AXPY = AX.XP = \frac{2\sqrt{3}}{3}.\frac{2}{3} = \frac{4\sqrt{3}}{9}$$
$$cos(angleXPA) = \frac{PX}{AP} = \frac{1}{2} \implies angle(XPY) = \frac{2\pi}{3}$$
Hence the area in question is
$$\frac{4\sqrt{3}}{9} - \frac{1}{2}(\frac{2}{3})^2\frac{2\pi}{3} = \frac{4\sqrt{3}}{9} - \frac{4\pi}{27}$$
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