The centres of two circles `C_(1) and C_(2)` each of unit radius are at a distance of 6 unit from each other. Let P be the mid-point of the line segment joining the centres of `C_(1) and C_(2)` and C be a circle touching circles `C_(1) and C_(2)` externally. If a common tangent to `C_(1)` and C passing through P is also a common tangent to `C_(2)` and C, then the radius of the circle C, is

Explanation

To solve the problem, we need to find the radius of circle \( C \) that touches the two circles \( C_1 \) and \( C_2 \) externally. Here are the steps to derive the solution:

### Step 1: Understand the Configuration
We have two circles \( C_1 \) and \( C_2 \) with centers \( A \) and \( B \) respectively. Both circles have a radius of 1 unit and are 6 units apart. The midpoint \( P \) of the line segment \( AB \) is given, and we need to find the radius \( r \) of circle \( C \) that touches both \( C_1 \) and \( C_2 \) externally.

### Step 2: Determine the Coordinates
Let’s place the circles in a coordinate system for easier calculations:
- Circle \( C_1 \) at \( A(0, 0) \)
- Circle \( C_2 \) at \( B(6, 0) \)

The midpoint \( P \) can be calculated as:
\[ P\left(\frac{0 + 6}{2}, 0\right) = P(3, 0) \]

### Step 3: Set Up the Geometry
The radius of circle \( C \) is \( r \). The distance from the center of circle \( C \) to the center of circle \( C_1 \) is \( 1 + r \) (since they touch externally), and the distance from the center of circle \( C \) to the center of circle \( C_2 \) is also \( 1 + r \).

### Step 4: Use the Right Triangle
Let the center of circle \( C \) be at point \( D \) located at \( (x, y) \). The distances can be expressed as:
\[ AD = \sqrt{x^2 + y^2} = 1 + r \]
\[ BD = \sqrt{(x - 6)^2 + y^2} = 1 + r \]

### Step 5: Set Up the Equations
From the above distances, we can set up the following equations:
1. \( x^2 + y^2 = (1 + r)^2 \)
2. \( (x - 6)^2 + y^2 = (1 + r)^2 \)

### Step 6: Expand and Simplify
Expanding the second equation:
\[ (x - 6)^2 + y^2 = (1 + r)^2 \]
\[ x^2 - 12x + 36 + y^2 = (1 + r)^2 \]

Now, substitute \( (1 + r)^2 \) from the first equation into the second:
\[ x^2 - 12x + 36 + y^2 = x^2 + y^2 \]
This simplifies to:
\[ -12x + 36 = 0 \]
Thus, we find:
\[ x = 3 \]

### Step 7: Substitute Back to Find \( y \)
Substituting \( x = 3 \) back into the first equation:
\[ 3^2 + y^2 = (1 + r)^2 \]
\[ 9 + y^2 = (1 + r)^2 \]

### Step 8: Use the Tangent Condition
Since \( DP \) is perpendicular to \( AB \), we can find \( y \) using the tangent condition:
Using the right triangle formed by \( P, D, \) and the tangent point, we can express:
\[ DP = r \]
\[ PB = 3 \quad \text{(half of the distance between the circles)} \]

Using the Pythagorean theorem in triangle \( PBD \):
\[ PB^2 + DP^2 = BD^2 \]
\[ 3^2 + r^2 = (1 + r)^2 \]
Expanding gives:
\[ 9 + r^2 = 1 + 2r + r^2 \]
This simplifies to:
\[ 9 = 1 + 2r \]
Thus,
\[ 2r = 8 \quad \Rightarrow \quad r = 4 \]

### Conclusion
The radius of circle \( C \) is \( r = 4 \) units.