If the circle `(x-6)^(2)+y^(2)=r^(2)` and the parabola `y^(2)=4x` have maximum number of common chords, then the least integral value of r is __________ .

Answer

To solve the problem, we need to determine the least integral value of \( r \) such that the circle \((x-6)^2 + y^2 = r^2\) and the parabola \(y^2 = 4x\) have the maximum number of common chords (which is 4 distinct intersection points).

### Step-by-Step Solution:

1. **Write the equations**:
- Circle: \((x - 6)^2 + y^2 = r^2\)
- Parabola: \(y^2 = 4x\)

2. **Substitute the parabola's equation into the circle's equation**:
- Replace \(y^2\) in the circle's equation with \(4x\):
\[ (x - 6)^2 + 4x = r^2 \]

3. **Expand the circle's equation**:
\[ (x - 6)^2 = x^2 - 12x + 36 \]
Thus, the equation becomes:
\[ x^2 - 12x + 36 + 4x = r^2 \]
Simplifying this, we have:
\[ x^2 - 8x + 36 - r^2 = 0 \]

4. **Identify the coefficients**:
- Here, \(a = 1\), \(b = -8\), and \(c = 36 - r^2\).

5. **Use the discriminant for intersection points**:
- For the circle and parabola to intersect at 4 distinct points, the discriminant of the quadratic equation must be positive:
\[ D = b^2 - 4ac > 0 \]
Substituting the values:
\[ D = (-8)^2 - 4 \cdot 1 \cdot (36 - r^2) > 0 \]
This simplifies to:
\[ 64 - 4(36 - r^2) > 0 \]

6. **Simplify the inequality**:
\[ 64 - 144 + 4r^2 > 0 \]
\[ 4r^2 - 80 > 0 \]
Dividing everything by 4:
\[ r^2 - 20 > 0 \]
Thus:
\[ r^2 > 20 \]

7. **Find the least integral value of \( r \)**:
- Taking the square root:
\[ r > \sqrt{20} = 2\sqrt{5} \approx 4.47 \]
The least integral value of \( r \) that satisfies this inequality is:
\[ r = 5 \]

### Final Answer:
The least integral value of \( r \) is **5**.