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 $\angle AOP$ is $60^\circ$, find the value of $\log_{e}(\frac{r}{s})$.

To solve this problem, let's go through it step-by-step using the information provided.

### Step 1: Set up the geometry and variables

1. **Circle centers**: We have two circles, one centered at $ O $ with radius $ r $ and the other centered at $ P $ with radius $ s $.
2. **Intersection points**: These circles intersect at points $ A $ and $ B $.
3. **Key points on the line segment $ OP $**:
   - The line segment $ OP $ has a length of $ 6 $.
   - $ OC = 2 $, so $ C $ is on the circle with center $ O $.
   
   Since $ OC = 2 $ is on the circle centered at $ O $ with radius $ r $, it follows that $ r = OC = 2 $.

### Step 2: Use cosine rule in $ \triangle OAP $

We are given that $ \angle AOP = 60^\circ $. Let $ OP = 6 $ and $ OA = r = 2 $. We aim to find the radius $ s = PB $.

Applying the **cosine rule** in $ \triangle OAP $:
\[
AP^2 = OA^2 + OP^2 - 2 \cdot OA \cdot OP \cdot \cos(\angle AOP)
\]
Substitute the values:
\[
AP^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos(60^\circ)
\]
Since $ \cos(60^\circ) = \frac{1}{2} $:
\[
AP^2 = 4 + 36 - 2 \cdot 2 \cdot 6 \cdot \frac{1}{2}
\]
\[
AP^2 = 40 - 12 = 28
\]
Thus,
\[
AP = \sqrt{28} = 2\sqrt{7}
\]

### Step 3: Recognize $ AP = s $

Since $ A $ is on both circles, the distance $ AP $ is the radius $ s $ of the circle centered at $ P $. Therefore:
\[
s = 2\sqrt{7}
\]

### Step 4: Find $ \log_e \left( \frac{r}{s} \right) $

Now we have:
- $ r = 2 $
- $ s = 2\sqrt{7} $

So,
\[
\frac{r}{s} = \frac{2}{2\sqrt{7}} = \frac{1}{\sqrt{7}} = 7^{-\frac{1}{2}}
\]
Thus,
\[
\log_e \left( \frac{r}{s} \right) = \log_e \left( 7^{-\frac{1}{2}} \right) = -\frac{1}{2} \log_e 7
\]

### Final Answer
\[
\log_e \left( \frac{r}{s} \right) = -\frac{1}{2} \log_e 7
\]

Would you like more details or have any questions?

Here are some related questions:
1. How would the answer change if $ \angle AOP $ were different?
2. What if $ OP $ were perpendicular to $ \angle AOP $ instead of $ 60^\circ $?
3. Can we generalize this approach for arbitrary values of $ OP $, $ \angle AOP $, $ OC $?
4. What are the applications of intersection points in geometry problems?
5. How does the cosine rule help in determining side lengths in triangles?

**Tip:** For problems with intersecting circles, visualize key segments and use geometric properties like cosine rule for calculations.

This problem explores intersecting circles, applying the cosine rule to find lengths and solving for a logarithmic ratio, log_e(r/s), based on given angles and distances. Ideal for students in Grades 10-12.

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