If `A(cos alpha, sin alpha), B(sin alpha, -cos alpha), C(1,2)` are the vertices of a triangle, then as `alpha` varies the locus of centroid of the `DeltaABC` is a circle whose radius is :

Options

A. `(2sqrt(2))/(3)`
B. `sqrt((4)/(3))`
C. `(2)/(3)`
D. `sqrt((2)/(9))`

Answer

To find the radius of the circle that represents the locus of the centroid of triangle ABC, we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of triangle ABC are given as:
- \( A(\cos \alpha, \sin \alpha) \)
- \( B(\sin \alpha, -\cos \alpha) \)
- \( C(1, 2) \)

### Step 2: Find the coordinates of the centroid
The centroid \( G(h, k) \) of triangle ABC can be calculated using the formula:
\[ G(h, k) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates of points A, B, and C:
\[ h = \frac{\cos \alpha + \sin \alpha + 1}{3} \]
\[ k = \frac{\sin \alpha - \cos \alpha + 2}{3} \]

### Step 3: Rearrange the equations for h and k
We can rearrange the equations for \( h \) and \( k \):
1. From \( h \):
\[ 3h = \cos \alpha + \sin \alpha + 1 \implies \cos \alpha + \sin \alpha = 3h - 1 \]
2. From \( k \):
\[ 3k = \sin \alpha - \cos \alpha + 2 \implies \sin \alpha - \cos \alpha = 3k - 2 \]

### Step 4: Square and add the two equations
Now we square both equations and add them:
\[ (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2 = (3h - 1)^2 + (3k - 2)^2 \]
Expanding both sides:
- Left side:
\[ \cos^2 \alpha + 2\cos \alpha \sin \alpha + \sin^2 \alpha + \sin^2 \alpha - 2\cos \alpha \sin \alpha + \cos^2 \alpha = 1 + 1 = 2 \]
- Right side:
\[ (3h - 1)^2 + (3k - 2)^2 \]

### Step 5: Set the equation
Now we have:
\[ 2 = (3h - 1)^2 + (3k - 2)^2 \]

### Step 6: Rearranging to the standard form of a circle
Rearranging gives:
\[ (3h - 1)^2 + (3k - 2)^2 = 2 \]
This represents a circle centered at \( (1/3, 2/3) \) with radius \( \sqrt{2} \).

### Step 7: Find the radius
The radius \( r \) of the circle can be found from the equation:
\[ r = \sqrt{2} \]

### Conclusion
Thus, the radius of the circle representing the locus of the centroid of triangle ABC is:
\[ \boxed{\sqrt{2}} \]