The line that is tangent to the circle $x^2+y^2=25$ at the point $\left(3,4\right)$ intersects the $x$-axis at $\left(k,0\right)$. What is $k$?

$\text{(A) }\frac{25}{4}\qquad\text{(B) }\frac{19}{3}\qquad\text{(C) }25\qquad\text{(D) }\frac{25}{3}\qquad\text{(E) }-\frac{7}{3}$
1. **Identify the given circle and the point of tangency:**
   The equation of the circle is \(x^2 + y^2 = 25\). This is a circle centered at the origin \((0,0)\) with radius 5. The point of tangency is \((3, 4)\).

2. **Determine the slope of the radius at the point of tangency:**
   The radius of the circle at the point \((3, 4)\) is the line segment from the origin \((0,0)\) to \((3, 4)\). The slope of this radius is:
   \[
   \text{slope of radius} = \frac{4 - 0}{3 - 0} = \frac{4}{3}
   \]

3. **Find the slope of the tangent line:**
   The tangent line at any point on a circle is perpendicular to the radius at that point. Therefore, the slope of the tangent line is the negative reciprocal of the slope of the radius:
   \[
   \text{slope of tangent line} = -\frac{1}{\left(\frac{4}{3}\right)} = -\frac{3}{4}
   \]

4. **Write the equation of the tangent line:**
   Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency \((3, 4)\):
   \[
   y - 4 = -\frac{3}{4}(x - 3)
   \]
   Simplify this equation:
   \[
   y - 4 = -\frac{3}{4}x + \frac{9}{4}
   \]
   \[
   y = -\frac{3}{4}x + \frac{9}{4} + 4
   \]
   \[
   y = -\frac{3}{4}x + \frac{9}{4} + \frac{16}{4}
   \]
   \[
   y = -\frac{3}{4}x + \frac{25}{4}
   \]

5. **Find the x-intercept of the tangent line:**
   The x-intercept occurs where \(y = 0\):
   \[
   0 = -\frac{3}{4}x + \frac{25}{4}
   \]
   Solving for \(x\):
   \[
   \frac{3}{4}x = \frac{25}{4}
   \]
   \[
   x = \frac{25}{4} \div \frac{3}{4} = \frac{25}{4} \times \frac{4}{3} = \frac{25}{3}
   \]

6. **Conclusion:**
   The x-intercept of the tangent line is \(\left(\frac{25}{3}, 0\right)\).

The final answer is \(\boxed{\frac{25}{3}}\)