There exists a fraction $x$ that satisfies $ \sqrt{x^2+5} - x = \tfrac{1}{3}$. What is the sum of the numerator and denominator of this fraction?

$\textbf{(A) }8\qquad\textbf{(B) }21\qquad\textbf{(C) }25\qquad\textbf{(D) }32\qquad\textbf{(E) }34$
1. Start with the given equation:
   \[
   \sqrt{x^2 + 5} - x = \frac{1}{3}
   \]

2. Rearrange the equation to isolate the square root term:
   \[
   \sqrt{x^2 + 5} = x + \frac{1}{3}
   \]

3. Square both sides to eliminate the square root:
   \[
   (\sqrt{x^2 + 5})^2 = \left(x + \frac{1}{3}\right)^2
   \]
   \[
   x^2 + 5 = x^2 + \frac{2}{3}x + \frac{1}{9}
   \]

4. Simplify the equation by subtracting \(x^2\) from both sides:
   \[
   5 = \frac{2}{3}x + \frac{1}{9}
   \]

5. Clear the fraction by multiplying every term by 9:
   \[
   45 = 6x + 1
   \]

6. Isolate \(x\) by subtracting 1 from both sides:
   \[
   44 = 6x
   \]

7. Solve for \(x\) by dividing both sides by 6:
   \[
   x = \frac{44}{6} = \frac{22}{3}
   \]

8. Verify that \(x = \frac{22}{3}\) satisfies the original equation:
   \[
   \sqrt{\left(\frac{22}{3}\right)^2 + 5} - \frac{22}{3} = \frac{1}{3}
   \]
   \[
   \sqrt{\frac{484}{9} + 5} - \frac{22}{3} = \frac{1}{3}
   \]
   \[
   \sqrt{\frac{484}{9} + \frac{45}{9}} - \frac{22}{3} = \frac{1}{3}
   \]
   \[
   \sqrt{\frac{529}{9}} - \frac{22}{3} = \frac{1}{3}
   \]
   \[
   \frac{23}{3} - \frac{22}{3} = \frac{1}{3}
   \]
   \[
   \frac{1}{3} = \frac{1}{3}
   \]

   The solution is verified.

9. The fraction \(x = \frac{22}{3}\) has a numerator of 22 and a denominator of 3. The sum of the numerator and denominator is:
   \[
   22 + 3 = 25
   \]

The final answer is \(\boxed{25}\).