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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?

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    $\begingroup$ It is better to divide the area into 2 right triangles (use Pythagoras) + a circular sector with area $\frac{\theta}{2} r^2$. $\endgroup$ – Jean Marie Feb 15 at 9:07
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    $\begingroup$ This is called a circle segment. mathopenref.com/segmentarea.html $\endgroup$ – Yves Daoust Feb 15 at 10:53
  • $\begingroup$ Could you provide an image? $\endgroup$ – Dr. Mathva Feb 15 at 11:32
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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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  • $\begingroup$ How come PX is 2/3? Isn't AX a chord? $\endgroup$ – suklay Feb 17 at 1:47
  • $\begingroup$ The question said that AX and AY are tangents to circle P therefore PX is a radius of length 2/3. $\endgroup$ – KY Tang Feb 17 at 19:08

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