If `cos alpha, cos beta and cos gamma `are the roots of the equation `9x ^(3)-9x ^(2) -x+1 =0, alpha, beta , gamma in [0,pi]` then the radius of the circle whose centre is `(sum alpha, sum cos alpha)` and passing through `(2 sin ^(-1) (tan pi//4), 4)` is :

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the roots of the cubic equation
The given cubic equation is:
\[ 9x^3 - 9x^2 - x + 1 = 0 \]

We will factor this equation. We can rearrange it as:
\[ 9x^3 - 9x^2 - x + 1 = 0 \]

We can try to factor by grouping:
\[ 9x^2(x - 1) - 1(x - 1) = 0 \]
This gives us:
\[ (x - 1)(9x^2 - 1) = 0 \]

Now we can set each factor to zero:
1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( 9x^2 - 1 = 0 \) → \( 9x^2 = 1 \) → \( x^2 = \frac{1}{9} \) → \( x = \pm \frac{1}{3} \)

Thus, the roots are:
\[ x = 1, \quad x = \frac{1}{3}, \quad x = -\frac{1}{3} \]

### Step 2: Identify the angles
Given that the roots correspond to \( \cos \alpha, \cos \beta, \cos \gamma \), we can assign:
- \( \cos \alpha = 1 \) → \( \alpha = 0 \)
- \( \cos \beta = \frac{1}{3} \) → \( \beta = \cos^{-1}\left(\frac{1}{3}\right) \)
- \( \cos \gamma = -\frac{1}{3} \) → \( \gamma = \cos^{-1}\left(-\frac{1}{3}\right) \)

### Step 3: Find the relationship between angles
From the properties of cosine, we know:
\[ \beta + \gamma = \pi \]
This is because \( \cos \beta = -\cos \gamma \).

### Step 4: Calculate the center of the circle
The center of the circle is given by:
\[ C = \left( \sum \alpha, \sum \cos \alpha \right) \]

Calculating \( \sum \alpha \):
\[ \sum \alpha = 0 + \cos^{-1}\left(\frac{1}{3}\right) + \cos^{-1}\left(-\frac{1}{3}\right) = \pi \]

Calculating \( \sum \cos \alpha \):
\[ \sum \cos \alpha = 1 + \frac{1}{3} - \frac{1}{3} = 1 \]

Thus, the center of the circle is:
\[ C = (\pi, 1) \]

### Step 5: Find the point through which the circle passes
The point is given as:
\[ P = \left( 2 \sin^{-1}(\tan \frac{\pi}{4}), 4 \right) \]

Since \( \tan \frac{\pi}{4} = 1 \):
\[ \sin^{-1}(1) = \frac{\pi}{2} \]
Thus:
\[ P = (2 \cdot \frac{\pi}{2}, 4) = (\pi, 4) \]

### Step 6: Calculate the radius of the circle
Using the distance formula, the radius \( r \) is the distance from the center \( C(\pi, 1) \) to the point \( P(\pi, 4) \):
\[ r = \sqrt{(\pi - \pi)^2 + (4 - 1)^2} = \sqrt{0 + 3^2} = \sqrt{9} = 3 \]

### Conclusion
The radius of the circle is:
\[ \boxed{3} \]