Finding Perimeter of Shape "Two circles of radii 5cm and 12cm overlap so that the distance between their centers is 13cm.
Find the perimeter of the shape."
This question was from a chapter about circle measure under the length of an arc in P1 AS Mathematics.
Thanks!
 A: The perimeter is the sum of the perimeters of the two circles less the lengths of arcs inside. So referring to the diagram below it is $2\pi(12+5)-2CD-2CE$. The arc $CD$ has length $12\angle ABC$ (where the angle is measured in radians) and the arc $CE$ has length $5\angle BAC$.
We have $\tan ABC=\frac{AC}{BC}=\frac{5}{12}$, so $\angle ABC=\tan^{-1}\frac{5}{12}$ and $\angle BAC=\frac{\pi}{2}-\tan^{-1}\frac{5}{12}$.
Thus the perimeter is $29\pi-14\tan^{-1}\frac{5}{12}\approx85.6$

A: Here is my answer. You have 2 circumferences: $x^2+y^2 = 25$ and $(x-13)^2 + y^2 = 144$. You can find the intersection between the two circumferences putting in a sistem the two equations. You find 2 points: $(1.92,4.62)$ and $(1.92,-4.62)$. Now you had to find the angle between center of both cirumferences and this point (i used a program). The angle is 225.24° for the first one, and 314.76° for the second one. Then use the formula Arch = (2*pirangle)/360 for both circumferences and add the results. You obtain $27.23*\pi$
