# 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?

• It is better to divide the area into 2 right triangles (use Pythagoras) + a circular sector with area $\frac{\theta}{2} r^2$. – Jean Marie Feb 15 at 9:07
• This is called a circle segment. mathopenref.com/segmentarea.html – Yves Daoust Feb 15 at 10:53
• Could you provide an image? – 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

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

• How come PX is 2/3? Isn't AX a chord? – suklay Feb 17 at 1:47
• The question said that AX and AY are tangents to circle P therefore PX is a radius of length 2/3. – KY Tang Feb 17 at 19:08