A circle passes through the points (2, 2) and (9, 9) and touches the x-axis. The absolute value of the difference of possible x-coordinate of the point of contact is______.

Explanation

To solve the problem step by step, we need to find the absolute value of the difference of possible x-coordinates of the point of contact of the circle with the x-axis. The circle passes through the points (2, 2) and (9, 9) and touches the x-axis.

### Step 1: Write the general equation of the circle
The general equation of a circle that passes through two points (x1, y1) and (x2, y2) can be expressed as:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) + \lambda (x - y) = 0 \]
For our points (2, 2) and (9, 9), we can substitute:
\[ (x - 2)(x - 9) + (y - 2)(y - 9) + \lambda (x - y) = 0 \]

### Step 2: Substitute y = 0 (since the circle touches the x-axis)
Since the circle touches the x-axis, we set \(y = 0\):
\[ (x - 2)(x - 9) + (0 - 2)(0 - 9) + \lambda (x - 0) = 0 \]
This simplifies to:
\[ (x - 2)(x - 9) + 18 + \lambda x = 0 \]

### Step 3: Expand the equation
Expanding the first term:
\[ x^2 - 11x + 18 + \lambda x = 0 \]
Combining like terms gives:
\[ x^2 + (-11 + \lambda)x + 18 = 0 \]

### Step 4: Condition for the circle to touch the x-axis
For the circle to touch the x-axis, the quadratic must have exactly one solution. This occurs when the discriminant is zero:
\[ b^2 - 4ac = 0 \]
Here, \(a = 1\), \(b = -11 + \lambda\), and \(c = 18\). Thus:
\[ (-11 + \lambda)^2 - 4(1)(18) = 0 \]
Expanding this gives:
\[ (-11 + \lambda)^2 - 72 = 0 \]

### Step 5: Solve the equation
Solving for \(\lambda\):
\[ (-11 + \lambda)^2 = 72 \]
Taking the square root:
\[ -11 + \lambda = \pm \sqrt{72} \]
Calculating \(\sqrt{72} = 6\sqrt{2}\):
\[ \lambda = 11 \pm 6\sqrt{2} \]

### Step 6: Substitute \(\lambda\) back to find x-coordinates
Now we substitute \(\lambda\) back into the equation:
1. For \(\lambda = 11 + 6\sqrt{2}\):
\[ x^2 + (-11 + (11 + 6\sqrt{2}))x + 18 = 0 \implies x^2 + 6\sqrt{2}x + 18 = 0 \]
The roots can be calculated using the quadratic formula.

2. For \(\lambda = 11 - 6\sqrt{2}\):
\[ x^2 + (-11 + (11 - 6\sqrt{2}))x + 18 = 0 \implies x^2 - 6\sqrt{2}x + 18 = 0 \]
The roots can also be calculated using the quadratic formula.

### Step 7: Calculate the x-coordinates
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
1. For \(\lambda = 11 + 6\sqrt{2}\):
\[ x = \frac{-6\sqrt{2} \pm 0}{2} = -3\sqrt{2} \]

2. For \(\lambda = 11 - 6\sqrt{2}\):
\[ x = \frac{6\sqrt{2} \pm 0}{2} = 3\sqrt{2} \]

### Step 8: Find the absolute value of the difference
The possible x-coordinates are \(3\sqrt{2}\) and \(-3\sqrt{2}\). The absolute value of the difference is:
\[ |3\sqrt{2} - (-3\sqrt{2})| = |3\sqrt{2} + 3\sqrt{2}| = |6\sqrt{2}| \]
Thus, the absolute value of the difference is \(6\sqrt{2}\).

### Final Answer
The absolute value of the difference of possible x-coordinates of the point of contact is \(6\sqrt{2}\).