We are tasked with finding the value of $\log_{e}\left(\frac{r}{s}\right)$, where $r$ and $s$ are the radii of two intersecting circles, one with center $O$ and radius $r$, and the other with center $P$ and radius $s$. The two circles intersect at points $A$ and $B$. We are given the following information:

• The length of the line segment $OP = 6$.
• The length of $OC = 2$, where $C$ is the point of intersection of line segment $OP$ with the circle centered at $O$.
• The angle $\angle AOP = 60^\circ$.

Step 1: Interpret the geometry of the problem

We can assume that $O$ is the center of the circle with radius $r$, and $P$ is the center of the circle with radius $s$. The line segment $OP$ is the distance between the centers of the circles, which is given as $6$.

The line segment $OC$, which is part of $OP$, intersects the circle with center $O$ at point $C$, and the length $OC = 2$. Hence, $C$ is on the circle with center $O$ and radius $r$.

Step 2: Use the cosine rule in $\triangle AOP$

Since the angle $\angle AOP = 60^\circ$ and we know the distance $OP = 6$, we can apply the cosine rule in triangle $\triangle AOP$ to express the relationship between $r$, $s$, and the distance $OP$.

The cosine rule states: $AP^2 = AO^2 + OP^2 - 2 \cdot AO \cdot OP \cdot \cos(\angle AOP)$ Substitute the known values:

• $AO = r$
• $OP = 6$
• $\cos(60^\circ) = \frac{1}{2}$

Thus, the equation becomes: $AP^2 = r^2 + 6^2 - 2 \cdot r \cdot 6 \cdot \frac{1}{2}$ $AP^2 = r^2 + 36 - 6r$ AP^2 = r^2 - 6r + 36 \tag{1}

Step 3: Equation for the second circle

Since $A$ lies on both circles, the distance $AP$ is also related to the radius $s$ of the second circle. For the circle centered at $P$, the distance from $P$ to $A$ is equal to the radius $s$, so we have: $AP = s$ From equation $(1)$, we can substitute $AP = s$ into the equation: s^2 = r^2 - 6r + 36 \tag{2}

Step 4: Use the information about point $C$

We are given that $OC = 2$. Since point $C$ lies on the circle centered at $O$ with radius $r$, we know: $OC = r = 2$

Step 5: Solve for $r$ and $s$

Substituting $r = 2$ into equation $(2)$: $s^2 = 2^2 - 6(2) + 36$ $s^2 = 4 - 12 + 36$ $s^2 = 28$ $s = \sqrt{28} = 2\sqrt{7}$

Step 6: Calculate $\log_{e}\left(\frac{r}{s}\right)$

Now that we have $r = 2$ and $s = 2\sqrt{7}$, we can compute the desired logarithmic value: $\log_{e}\left(\frac{r}{s}\right) = \log_{e}\left(\frac{2}{2\sqrt{7}}\right) = \log_{e}\left(\frac{1}{\sqrt{7}}\right)$ $\log_{e}\left(\frac{1}{\sqrt{7}}\right) = -\frac{1}{2} \log_{e}(7)$

Thus, the value of $\log_{e}\left(\frac{r}{s}\right)$ is $-\frac{1}{2} \log_{e}(7)$.

Conclusion:

The value of $\log_{e}\left(\frac{r}{s}\right)$ is $-\frac{1}{2} \log_{e}(7)$.

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Would you like further clarification or more details on any part of this solution?

Related questions:

1. How does the cosine rule help in solving triangle-related problems in geometry?
2. Why is the logarithm of a fraction related to the difference of logarithms?
3. What are the general properties of intersecting circles in geometry?
4. How can the angle between intersecting circles affect the relationship between their radii?
5. What other trigonometric properties could be useful in circle geometry?

Tip:

When solving geometry problems involving circles and angles, the cosine rule is a powerful tool to relate sides and angles in non-right triangles.