Hello, castle!

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 $\displaystyle P$ be the center of the 12-radius circle,

. . and $\displaystyle Q$ be the center of the 9-radius circle.

Now look at $\displaystyle \Delta APQ.$

Code:

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

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

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

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

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

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

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

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

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

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

Therefore, the external circumference is:

. . $\displaystyle 59.95419707 + 39.85735388 \;=\;99.81155095$