If `cos alpha, cos beta and c 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 given problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Identify the roots of the cubic equation
The roots of the equation \( 9x^3 - 9x^2 - x + 1 = 0 \) are \( \cos \alpha, \cos \beta, \cos \gamma \).

### Step 2: Factor the cubic equation
We can factor the cubic equation:
\[ 9x^3 - 9x^2 - x + 1 = 0 \]
We can try to find rational roots using the Rational Root Theorem. Testing \( x = 1 \):
\[ 9(1)^3 - 9(1)^2 - (1) + 1 = 9 - 9 - 1 + 1 = 0 \]
So, \( x = 1 \) is a root. We can factor out \( (x - 1) \):
\[ 9x^3 - 9x^2 - x + 1 = (x - 1)(9x^2 - 1) \]
Now, we can factor \( 9x^2 - 1 \):
\[ 9x^2 - 1 = (3x - 1)(3x + 1) \]
Thus, the complete factorization is:
\[ 9x^3 - 9x^2 - x + 1 = (x - 1)(3x - 1)(3x + 1) \]

### Step 3: Find the roots
Setting each factor to zero gives us:
1. \( x - 1 = 0 \) → \( x = 1 \) → \( \cos \alpha = 1 \) → \( \alpha = 0 \)
2. \( 3x - 1 = 0 \) → \( x = \frac{1}{3} \) → \( \cos \beta = \frac{1}{3} \) → \( \beta = \cos^{-1}\left(\frac{1}{3}\right) \)
3. \( 3x + 1 = 0 \) → \( x = -\frac{1}{3} \) → \( \cos \gamma = -\frac{1}{3} \) → \( \gamma = \cos^{-1}\left(-\frac{1}{3}\right) \)

### Step 4: Calculate \( \sum \alpha \) and \( \sum \cos \alpha \)
Using the properties of roots:
- \( \alpha + \beta + \gamma = \frac{9}{9} = 1 \) (from Vieta's formulas)
- \( \cos \alpha + \cos \beta + \cos \gamma = \frac{9}{9} = 1 \)

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

### Step 5: Center of the circle
The center of the circle is:
\[ \left(\sum \alpha, \sum \cos \alpha\right) = \left(\pi, 1\right) \]

### Step 6: Find the point through which the circle passes
The point through which the circle passes is given as:
\[ \left(2 \sin^{-1}(\tan(\frac{\pi}{4})), 4\right) \]
Since \( \tan(\frac{\pi}{4}) = 1 \):
\[ \sin^{-1}(1) = \frac{\pi}{2} \quad \Rightarrow \quad 2 \sin^{-1}(1) = 2 \cdot \frac{\pi}{2} = \pi \]
Thus, the point is \( (\pi, 4) \).

### Step 7: Calculate the radius of the circle
Using the distance formula to find the radius \( r \):
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Where:
- \( (x_1, y_1) = (\pi, 1) \)
- \( (x_2, y_2) = (\pi, 4) \)

Calculating the distance:
\[ r = \sqrt{(\pi - \pi)^2 + (4 - 1)^2} = \sqrt{0 + 3^2} = \sqrt{9} = 3 \]

### Final Answer
The radius of the circle is \( \boxed{3} \).