# Finding the area of the intersection of two circles in terms of distance between centres of circles

I study maths as a hobby and have come across this problem. Two circles with centres A and B intersect at points P and Q, such that $$\angle APB$$ is a right angle. If AB = xcm and $$\angle PAQ = \frac{1}{3}\pi$$ radians, find in terms of x the length of the perimeter and the area of the region common to the two circles.

I calculate the area of the segment APQ to be $$\frac{\pi}{6}r^2$$, where r = radius. The area of $$\triangle APQ$$ I calculate as $$\frac{1}{2}r^2\sin \frac{\pi}{3} = \frac{1}{2}r^2\sqrt3$$

I also know that to find the right hand side of the central segment I need to subtract the area of the triangle APQ from the area of the segment APQ.

But I cannot proceed any further and certainly not in terms of the length x.

This is the diagram as I visualise it: ## 2 Answers

HINT

We know that $$\angle PAB=\frac{\pi}{6}$$, hence the radius of the larger circle is $$x\cos{\frac{\pi}{6}}=\frac{x\sqrt3}{2}$$. Similarly, the radius of the smaller circle is $$x\sin{\frac{\pi}{6}}=\frac{x}{2}$$.

Does that help? If you need any more help, please don't hesitate to ask. I have a full solution ready to post if you need it.

• Thank you so much. – Steblo Dec 26 '20 at 11:34
• @Steblo you are most welcome, I'm very glad to have helped you! :) Feel free to ask any more questions. – A-Level Student Dec 26 '20 at 19:49

Hint. For the common area, $$\text{area of sector PBQ}+\text{area of sector PAQ}=\text{area of }\square\text{PBQA}+\text{common area of circles}$$

where \begin{align*}&\text{area of sector PAQ}=\frac16\pi r_1^2\\ &\text{area of sector PBQ}=\frac13\pi r_2^2\\ &\text{area of }\square\text{PBQA}=AB\times PD=xr_1\sin\angle PAD\end{align*}

Can you find $$r_1,r_2$$ in terms of $$x$$? Use the fact that $$\triangle APB$$ is right-angled.

• What do you mean by $\text{area of }\square PBQ$? What square is there? – A-Level Student Dec 24 '20 at 16:28
• @A-LevelStudent I meant quadrilateral PBQA, fixed now. – Shubham Johri Dec 24 '20 at 17:07
• Ah, thank you :) – A-Level Student Dec 24 '20 at 17:09