Find the mean value of the function $\ln r$ on the circle $(x - a)^2 + (y-b)^2 = R^2$ (of the function $1/r$ on the sphere).
To find the mean value of the function \( \frac{1}{r} \) on the sphere \( (x-a)^2 + (y-b)^2 + (z-c)^2 = R^2 \), we need to integrate the function over the surface of the sphere and then divide by the surface area of the sphere.

1. **Surface Area of the Sphere**:
   The surface area \( S \) of a sphere with radius \( R \) is given by:
   \[
   S = 4\pi R^2
   \]

2. **Parametrization of the Sphere**:
   The sphere can be parametrized using spherical coordinates:
   \[
   \begin{cases}
   x = a + R \cos(t) \cos(u) \\
   y = b + R \sin(t) \cos(u) \\
   z = c + R \sin(u)
   \end{cases}
   \]
   where \( t \in [0, 2\pi] \) and \( u \in [-\pi/2, \pi/2] \).

3. **Element of Surface Area**:
   The element of surface area \( dS \) in spherical coordinates is:
   \[
   dS = R^2 \cos(u) \, du \, dt
   \]

4. **Integrand**:
   The function we need to integrate is \( \frac{1}{r} \), where \( r = \sqrt{x^2 + y^2 + z^2} \). Substituting the parametrization into \( r \), we get:
   \[
   r = \sqrt{(a + R \cos(t) \cos(u))^2 + (b + R \sin(t) \cos(u))^2 + (c + R \sin(u))^2}
   \]

5. **Integral Setup**:
   The integral to find the mean value of \( \frac{1}{r} \) over the sphere is:
   \[
   \frac{1}{S} \int_{S} \frac{1}{r} \, dS = \frac{1}{4\pi R^2} \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} \frac{1}{\sqrt{(a + R \cos(t) \cos(u))^2 + (b + R \sin(t) \cos(u))^2 + (c + R \sin(u))^2}} R^2 \cos(u) \, du \, dt
   \]

6. **Simplification**:
   Simplifying the integrand, we get:
   \[
   \frac{1}{4\pi R^2} \int_{-\pi/2}^{\pi/2} \int_{0}^{2\pi} \frac{R^2 \cos(u)}{\sqrt{a^2 + b^2 + c^2 + 2R \cos(u) (a \cos(t) + b \sin(t)) + 2cR \sin(u) + R^2}} \, du \, dt
   \]

7. **Evaluation**:
   This integral is quite complex and typically requires numerical methods or advanced techniques to evaluate. However, for a sphere centered at the origin (i.e., \( a = b = c = 0 \)), the integral simplifies significantly. In this case, \( r = R \), and the mean value of \( \frac{1}{r} \) over the sphere is simply:
   \[
   \frac{1}{R}
   \]

The final answer is \( \boxed{ \frac{1}{R} } \)