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?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

$$\frac{4\sqrt{3}}{9} - \frac{1}{2}(\frac{2}{3})^2\frac{2\pi}{3} = \frac{4\sqrt{3}}{9} - \frac{4\pi}{27}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.