# Finding external Circumference of 2 overlapping circles

• Jun 26th 2011, 09:09 AM
castle
Finding external Circumference of 2 overlapping circles
I marked the top of the intersection (where the 2 circles meet) "a" and the bottom "b". Then I made a triangle with the midpoint of AB and the centre of the larger circle. I used Sine Law to find the angles and then found the sector length. My final answer came to be 98.89. However my textbook has a different answer, am I doing something wrong here.
• Jun 26th 2011, 09:37 AM
Prove It
Re: Finding external Circumference of 2 overlapping circles
If you make isosceles triangles using the radii and the length AB you should have a 12, 12, 15 triangle, and a 9, 9. 15 triangle.

Since in each case you have three sides and an unknown angle, I would use the cosine rule to find the unknown angle.

Triangle 1:

\displaystyle \begin{align*} \cos{\theta} &= \frac{12^2 + 12^2 - 15^2}{2\cdot 12 \cdot 12} \\ \cos{\theta} &= \frac{144 + 144 - 225}{288} \\ \cos{\theta} &= \frac{63}{288} \\ \cos{\theta} &= \frac{7}{32} \\ \theta &= \arccos{\frac{7}{32}} \\ \theta &\approx 77.36^{\circ}\end{align*}

So the length of the major segment for Triangle 1 is $\displaystyle \frac{360 - \arccos{\frac{7}{32}}}{360} \cdot 2\pi \cdot 12 \approx 59.12\,\textrm{cm}$.

Now do the same for the second circle and triangle.
• Jun 26th 2011, 10:08 AM
Soroban
Re: Finding external Circumference of 2 overlapping circles
Hello, castle!

Quote:

I marked the top of the intersection (where the 2 circles meet) "A" and the bottom "B".
Then I made a triangle with the midpoint of AB and the centre of the larger circle.
I used Sine Law to find the angles and then found the sector length.
My final answer came to be 98.89.
However my textbook has a different answer. . What is it?

Let $P$ be the center of the 12-radius circle,
. . and $Q$ be the center of the 9-radius circle.

Now look at $\Delta APQ.$

Code:

A
*
*  *
12  *    *  9
*        *
*          *
* α          β *
P *  *  *  *  *  *  * Q
15

Note that we have a 3-4-5 right triangle.

Hence: . $\cos\alpha \:=\:\frac{12}{15} \quad\Rightarrow\quad \alpha \:=\:\cos^{-1}(0.8) \:=\:0.643501109\text{ (radians)}$

Then: . $\angle APB \:=\:2\alpha \:=\:1.287002218$

And: $\text{major }\angle APB \:=\:2\pi - 2\alpha \:=\:4.996183089$

Hence: $\text{major arc }\overline{APB} \:=\:12(2\alpha) \:=\:59.95419707$

We have: . $\beta \:=\:\tfrac{\pi}{2} - \alpha \:=\:0.927295216$

Then: . $\angle AQB \:=\:2\beta \:=\:1.854590432$

And: $\text{major }\angle AQB \:=\:2\pi - 2\beta \:=\:4.428594875$

Hence: $\text{major arc }\overline{AQB} \:=\:9(2\beta) \:=\:39.85735388$

Therefore, the external circumference is:
. . $59.95419707 + 39.85735388 \;=\;99.81155095$

• Jun 26th 2011, 03:28 PM
castle
Re: Finding external Circumference of 2 overlapping circles
The textbook's answer is 102 cm, but I believe it may have just been a matter of rounding as my answer and Soroban's are very close to the textbook's.