Consider circles `C_(1) and C_(2)` touching both the axes and passing through `(4, 4),` then the x - intercept of the common chord of the circles is

Answer

To solve the problem, we need to find the x-intercept of the common chord of two circles \( C_1 \) and \( C_2 \) that touch both axes and pass through the point \( (4, 4) \).

### Step-by-Step Solution:

1. **Understanding the Circles**:
Since both circles touch the axes, their centers will be at \( (r, r) \) where \( r \) is the radius of the respective circles. Thus, the equations of the circles can be written as:
\[ C_1: (x - r_1)^2 + (y - r_1)^2 = r_1^2 \]
\[ C_2: (x - r_2)^2 + (y - r_2)^2 = r_2^2 \]
where \( r_1 \) and \( r_2 \) are the radii of circles \( C_1 \) and \( C_2 \) respectively.

2. **Substituting the Point (4, 4)**:
Since both circles pass through the point \( (4, 4) \), we can substitute \( x = 4 \) and \( y = 4 \) into the equations of the circles:
\[ (4 - r_1)^2 + (4 - r_1)^2 = r_1^2 \]
Simplifying this gives:
\[ 2(4 - r_1)^2 = r_1^2 \]
Expanding and rearranging:
\[ 2(16 - 8r_1 + r_1^2) = r_1^2 \]
\[ 32 - 16r_1 + 2r_1^2 = r_1^2 \]
\[ r_1^2 - 16r_1 + 32 = 0 \]

3. **Finding the Roots**:
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r_1 = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \]
\[ r_1 = \frac{16 \pm \sqrt{256 - 128}}{2} \]
\[ r_1 = \frac{16 \pm \sqrt{128}}{2} \]
\[ r_1 = \frac{16 \pm 8\sqrt{2}}{2} \]
\[ r_1 = 8 \pm 4\sqrt{2} \]
Thus, the radii are \( r_1 = 8 + 4\sqrt{2} \) and \( r_2 = 8 - 4\sqrt{2} \).

4. **Finding the Equation of the Common Chord**:
The equation of the common chord can be derived from the two circle equations. The common chord's equation is given by:
\[ 2x + 2y - (r_1 + r_2) = 0 \]
We already calculated \( r_1 + r_2 = 16 \). Thus:
\[ 2x + 2y - 16 = 0 \]
Simplifying gives:
\[ x + y = 8 \]

5. **Finding the X-Intercept**:
To find the x-intercept, set \( y = 0 \):
\[ x + 0 = 8 \implies x = 8 \]

### Final Answer:
The x-intercept of the common chord of the circles is \( \boxed{8} \).