A circle with center 
𝑂
O and radius 
𝑟
r intersects another circle with center 
𝑃
P and radius 
𝑠
s at two points 
𝐴
A and 
𝐵
B. The line segment 
𝑂
𝑃
OP intersects the circle with center 
𝑂
O at point 
𝐶
C. If the length of 
𝑂
𝑃
OP is 6, the length of 
𝑂
𝐶
OC is 2, and the angle 
∠
𝐴
𝑂
𝑃
∠AOP is 
6
0
∘
60 
∘
 , find the value of 
log
⁡
𝑒
(
𝑟
𝑠
)
log 
e
​
 ( 
s
r
​
 ).

Given:

- \( OP = 6 \)
- \( OC = 2 \)
- \( \angle AOP = 60^\circ \)

We need to find \( \log_e \left( \frac{s}{r} \right) \).

### Solution:

1. **Set Up Triangle \( \triangle OCP \):**
   Since \( C \) is a point on the circle with center \( O \), \( OC \) is a radius of this circle. Therefore, \( r = OC = 2 \).

2. **Applying the Cosine Rule in \( \triangle OCP \):**
   We know:
   - \( OP = 6 \)
   - \( \angle OCP = 60^\circ \)
   - \( OC = 2 \)

   Let \( CP = s \). Using the cosine rule in \( \triangle OCP \):
   \[
   s^2 = r^2 + OP^2 - 2 \cdot r \cdot OP \cdot \cos(60^\circ)
   \]

3. **Plugging in Values:**
   Substitute \( r = 2 \), \( OP = 6 \), and \( \cos(60^\circ) = \frac{1}{2} \):
   \[
   s^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \frac{1}{2}
   \]
   Simplify each term:
   \[
   s^2 = 4 + 36 - 2 \cdot 2 \cdot 6 \cdot \frac{1}{2}
   \]
  

A circle with center O and radius r intersects another circle with center P and radius s at two points A and B. The line segment OP intersects the circle with center O at point C. If the length of OP is 6, the length of OC is 2, and the angle ∠AOP is 60°, find the value of log_e(r/s).

This problem explores geometry and trigonometry concepts to determine log_e(r/s) in an intersecting circle scenario, given lengths OP=6, OC=2, and ∠AOP=60°. Detailed steps include using the cosine rule in triangle OCP to solve for radius ratios.

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