Bill draws two circles which intersect at $X,Y$. Let $P$ be the intersection of the common tangents to the two circles and let $Q$ be a point on the line segment connecting the centers of the two circles such that lines $PX$ and $QX$ are perpendicular. Given that the radii of the two circles are $3,4$ and the distance between the centers of these two circles is $5$, then the largest distance from $Q$ to any point on either of the circles can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $100m+n$.

[i]Proposed by Tristan Shin[/i]
1. **Define the centers and radii of the circles:**
   Let \( O_1 \) be the center of the smaller circle with radius \( r_1 = 3 \), and \( O_2 \) be the center of the larger circle with radius \( r_2 = 4 \). The distance between the centers \( O_1 \) and \( O_2 \) is given as \( 5 \).

2. **Determine the coordinates of \( P \):**
   The point \( P \) is the intersection of the common tangents to the two circles. Since the tangents are external, the distance from \( P \) to the points of tangency on each circle is equal to the radius of the respective circle. Let \( T_1 \) and \( T_2 \) be the points of tangency on the smaller and larger circles, respectively. By similar triangles, we have:
   \[
   \frac{PO_1}{PO_2} = \frac{PT_1}{PT_2} = \frac{3}{4}
   \]
   Since \( PO_2 - PO_1 = 5 \), let \( PO_1 = 3k \) and \( PO_2 = 4k \). Then:
   \[
   4k - 3k = 5 \implies k = 5
   \]
   Thus:
   \[
   PO_1 = 3 \cdot 5 = 15 \quad \text{and} \quad PO_2 = 4 \cdot 5 = 20
   \]
   Therefore, \( P \) is at \( (0, -15) \).

3. **Find the coordinates of \( X \):**
   Since \( O_1XO_2 \) forms a right triangle with sides 3, 4, and 5, we can place \( X \) using the ratio of the sides:
   \[
   X = 3 \left( \frac{3}{5}, \frac{4}{5} \right) = \left( \frac{9}{5}, \frac{12}{5} \right)
   \]

4. **Determine the slope of \( PX \):**
   The slope of \( PX \) is:
   \[
   \text{slope of } PX = \frac{\frac{12}{5} - (-15)}{\frac{9}{5} - 0} = \frac{\frac{12}{5} + 15}{\frac{9}{5}} = \frac{\frac{12 + 75}{5}}{\frac{9}{5}} = \frac{87}{9} = \frac{29}{3}
   \]

5. **Find the coordinates of \( Q \):**
   Since \( QX \) is perpendicular to \( PX \), the slope of \( QX \) is the negative reciprocal of \( \frac{29}{3} \), which is \( -\frac{3}{29} \). The equation of the line \( QX \) passing through \( X \left( \frac{9}{5}, \frac{12}{5} \right) \) is:
   \[
   y - \frac{12}{5} = -\frac{3}{29} \left( x - \frac{9}{5} \right)
   \]
   Since \( Q \) lies on the line segment connecting \( O_1 \) and \( O_2 \), which is the x-axis, \( Q \) has coordinates \( \left( \frac{15}{7}, 0 \right) \).

6. **Calculate the maximum distance from \( Q \) to any point on either circle:**
   The maximum distance from \( Q \) to any point on the smaller circle is the distance from \( Q \) to the farthest point on the circle, which is the diameter. The farthest point on the smaller circle from \( Q \) is \( (-3, 0) \):
   \[
   QR = \sqrt{\left( \frac{15}{7} + 3 \right)^2 + 0^2} = \sqrt{\left( \frac{15}{7} + \frac{21}{7} \right)^2} = \sqrt{\left( \frac{36}{7} \right)^2} = \frac{36}{7}
   \]
   The maximum distance from \( Q \) to any point on the larger circle is the distance from \( Q \) to the farthest point on the circle, which is the diameter. The farthest point on the larger circle from \( Q \) is \( (9, 0) \):
   \[
   QR = \sqrt{\left( 9 - \frac{15}{7} \right)^2 + 0^2} = \sqrt{\left( \frac{63}{7} - \frac{15}{7} \right)^2} = \sqrt{\left( \frac{48}{7} \right)^2} = \frac{48}{7}
   \]

7. **Determine the largest distance:**
   The largest distance is \( \frac{48}{7} \).

The final answer is \( \boxed{ 100m + n = 100 \cdot 48 + 7 = 4807 } \).