Which point on the circle $ (x \minus{} 11)^2 \plus{} (y \minus{} 13)^2 \equal{} 116$ is farthest from the point $ (41, 25)$?
1. **Identify the center and radius of the circle:**
   The given equation of the circle is \((x - 11)^2 + (y - 13)^2 = 116\). This is a standard form of a circle equation \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
   - Center: \((11, 13)\)
   - Radius: \(\sqrt{116}\)

2. **Find the equation of the line passing through the center of the circle and the point \((41, 25)\):**
   The slope \(m\) of the line passing through \((11, 13)\) and \((41, 25)\) is:
   \[
   m = \frac{25 - 13}{41 - 11} = \frac{12}{30} = \frac{2}{5}
   \]
   Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\), we get:
   \[
   y - 13 = \frac{2}{5}(x - 11)
   \]
   Simplifying, we get:
   \[
   y = \frac{2}{5}x + \frac{2}{5} \cdot (-11) + 13 = \frac{2}{5}x - \frac{22}{5} + 13 = \frac{2}{5}x + \frac{43}{5}
   \]

3. **Substitute the line equation into the circle equation:**
   Substitute \(y = \frac{2}{5}x + \frac{43}{5}\) into \((x - 11)^2 + (y - 13)^2 = 116\):
   \[
   (x - 11)^2 + \left(\frac{2}{5}x + \frac{43}{5} - 13\right)^2 = 116
   \]
   Simplify the expression inside the square:
   \[
   \frac{2}{5}x + \frac{43}{5} - 13 = \frac{2}{5}x + \frac{43 - 65}{5} = \frac{2}{5}x - \frac{22}{5}
   \]
   Thus, the equation becomes:
   \[
   (x - 11)^2 + \left(\frac{2}{5}x - \frac{22}{5}\right)^2 = 116
   \]

4. **Simplify and solve for \(x\):**
   \[
   (x - 11)^2 + \left(\frac{2}{5}x - \frac{22}{5}\right)^2 = 116
   \]
   \[
   (x - 11)^2 + \frac{4}{25}(x - 11)^2 = 116
   \]
   \[
   (x - 11)^2 \left(1 + \frac{4}{25}\right) = 116
   \]
   \[
   (x - 11)^2 \cdot \frac{29}{25} = 116
   \]
   \[
   (x - 11)^2 = 116 \cdot \frac{25}{29} = 100
   \]
   \[
   x - 11 = \pm 10
   \]
   Therefore, \(x = 21\) or \(x = 1\).

5. **Find the corresponding \(y\) values:**
   For \(x = 21\):
   \[
   y = \frac{2}{5} \cdot 21 + \frac{43}{5} = \frac{42}{5} + \frac{43}{5} = \frac{85}{5} = 17
   \]
   For \(x = 1\):
   \[
   y = \frac{2}{5} \cdot 1 + \frac{43}{5} = \frac{2}{5} + \frac{43}{5} = \frac{45}{5} = 9
   \]

6. **Determine which point is farthest from \((41, 25)\):**
   Calculate the distances:
   \[
   \text{Distance from } (21, 17) \text{ to } (41, 25) = \sqrt{(41 - 21)^2 + (25 - 17)^2} = \sqrt{20^2 + 8^2} = \sqrt{400 + 64} = \sqrt{464} = 2\sqrt{116}
   \]
   \[
   \text{Distance from } (1, 9) \text{ to } (41, 25) = \sqrt{(41 - 1)^2 + (25 - 9)^2} = \sqrt{40^2 + 16^2} = \sqrt{1600 + 256} = \sqrt{1856} = 4\sqrt{116}
   \]
   Since \(4\sqrt{116} > 2\sqrt{116}\), the point \((1, 9)\) is farther from \((41, 25)\).

The final answer is \(\boxed{(1, 9)}\).