The focal chord of `y^(2)=16x` is tangent to `(x-6)^(2)+y^(2)=2`.
Then the possible value of the square of slope of this chord is __________ .

Explanation

To solve the problem, we need to find the possible value of the square of the slope of the focal chord of the parabola \( y^2 = 16x \) that is tangent to the circle \( (x - 6)^2 + y^2 = 2 \).

### Step-by-Step Solution:

1. **Identify the Parabola and its Parameters:**
The given parabola is \( y^2 = 16x \). This can be rewritten in the standard form \( y^2 = 4ax \), where \( 4a = 16 \). Thus, we find \( a = 4 \). The focus of the parabola is at the point \( (4, 0) \).

**Hint:** Recall that the focus of a parabola in the form \( y^2 = 4ax \) is located at \( (a, 0) \).

2. **Equation of the Focal Chord:**
Let the slope of the focal chord be \( m \). The equation of the focal chord can be expressed as:
\[ y = mx - 4 \]

**Hint:** The equation of a line with slope \( m \) passing through the point \( (4, 0) \) can be derived using the point-slope form of a line.

3. **Identify the Circle and its Properties:**
The equation of the circle is given by \( (x - 6)^2 + y^2 = 2 \). From this, we can identify that the center of the circle is at \( (6, 0) \) and the radius is \( \sqrt{2} \).

**Hint:** The standard form of a circle is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius.

4. **Distance from the Center of the Circle to the Focal Chord:**
We need to find the distance from the center of the circle \( (6, 0) \) to the line \( y = mx - 4 \). The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( y = mx - 4 \), we can rewrite it as \( mx - y - 4 = 0 \) (where \( A = m, B = -1, C = -4 \)). Thus, the distance from \( (6, 0) \) to the line is:
\[ d = \frac{|m(6) - 0 - 4|}{\sqrt{m^2 + 1}} = \frac{|6m - 4|}{\sqrt{m^2 + 1}} \]

**Hint:** Use the formula for the distance from a point to a line to compute the distance.

5. **Set the Distance Equal to the Radius:**
Since the focal chord is tangent to the circle, this distance must equal the radius \( \sqrt{2} \):
\[ \frac{|6m - 4|}{\sqrt{m^2 + 1}} = \sqrt{2} \]

**Hint:** Set up the equation based on the condition of tangency.

6. **Clear the Denominator:**
Multiply both sides by \( \sqrt{m^2 + 1} \):
\[ |6m - 4| = \sqrt{2(m^2 + 1)} \]

**Hint:** Isolate the absolute value to simplify the equation.

7. **Square Both Sides:**
To eliminate the square root, square both sides:
\[ (6m - 4)^2 = 2(m^2 + 1) \]

**Hint:** Squaring both sides will remove the square root but be careful with the absolute value.

8. **Expand and Rearrange:**
Expanding the left side:
\[ 36m^2 - 48m + 16 = 2m^2 + 2 \]
Rearranging gives:
\[ 34m^2 - 48m + 14 = 0 \]

**Hint:** Combine like terms to form a standard quadratic equation.

9. **Solve the Quadratic Equation:**
Use the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 34, b = -48, c = 14 \):
\[ m = \frac{48 \pm \sqrt{(-48)^2 - 4 \cdot 34 \cdot 14}}{2 \cdot 34} \]

**Hint:** Calculate the discriminant to find the values of \( m \).

10. **Calculate the Discriminant:**
\[ (-48)^2 - 4 \cdot 34 \cdot 14 = 2304 - 1904 = 400 \]
Thus,
\[ m = \frac{48 \pm 20}{68} \]
This gives two possible values for \( m \):
\[ m_1 = 1 \quad \text{and} \quad m_2 = \frac{4}{17} \]

**Hint:** Solve for \( m \) using the quadratic formula.

11. **Find the Square of the Slope:**
The possible values of the square of the slope \( m^2 \) are:
\[ m_1^2 = 1^2 = 1 \quad \text{and} \quad m_2^2 = \left(\frac{4}{17}\right)^2 = \frac{16}{289} \]

**Hint:** Square each value of \( m \) to find \( m^2 \).

12. **Conclusion:**
The possible value of the square of the slope of the focal chord is:
\[ \boxed{1} \]