If `A(cosalpha, sinalpha),B(sinalpha- cosalpha) , C (2,1)` are the vertices of a `DeltaABC` , then as `alpha` varies the locus of its centroid is

Options

A. `x^2+y^2-2x-4y+1=0`
B. `3(x^2+y^2)-2x-4y+1=0`
C. `x^2+y^2-2x-4y+3=0`
D. none of these

Answer

To find the locus of the centroid of triangle ABC with vertices A(cosα, sinα), B(sinα - cosα, cosα), and C(2, 1) as α varies, we will follow these steps:

### Step 1: Determine the coordinates of the centroid G
The centroid G of a triangle with vertices A(x1, y1), B(x2, y2), and C(x3, y3) is given by the formula:
\[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

For our triangle ABC:
- A = (cosα, sinα)
- B = (sinα - cosα, cosα)
- C = (2, 1)

### Step 2: Calculate the x-coordinate of the centroid
Using the coordinates of A, B, and C:
\[ x_G = \frac{cosα + (sinα - cosα) + 2}{3} \]
Simplifying this:
\[ x_G = \frac{sinα + 2}{3} \]

### Step 3: Calculate the y-coordinate of the centroid
Now, we calculate the y-coordinate:
\[ y_G = \frac{sinα + cosα + 1}{3} \]

### Step 4: Set up the equations for x and y
We have:
\[ x = \frac{sinα + 2}{3} \quad \text{(1)} \]
\[ y = \frac{sinα + cosα + 1}{3} \quad \text{(2)} \]

### Step 5: Express sinα in terms of x
From equation (1):
\[ sinα = 3x - 2 \quad \text{(3)} \]

### Step 6: Substitute sinα into equation (2)
Substituting (3) into (2):
\[ y = \frac{(3x - 2) + cosα + 1}{3} \]
This simplifies to:
\[ y = \frac{3x - 1 + cosα}{3} \]
Thus:
\[ cosα = 3y - 3x + 1 \quad \text{(4)} \]

### Step 7: Use the identity sin²α + cos²α = 1
Using the Pythagorean identity:
\[ (3x - 2)^2 + (3y - 3x + 1)^2 = 1 \]

### Step 8: Expand and simplify
Expanding this:
1. For \( (3x - 2)^2 = 9x^2 - 12x + 4 \)
2. For \( (3y - 3x + 1)^2 = 9y^2 - 18xy + 9x^2 - 6y + 6x + 1 \)

Combining these:
\[ 9x^2 - 12x + 4 + 9y^2 - 18xy + 9x^2 - 6y + 6x + 1 = 1 \]
This simplifies to:
\[ 18x^2 + 9y^2 - 18xy - 6y - 6x + 4 = 0 \]

### Step 9: Rearranging the equation
Rearranging gives us:
\[ 18x^2 + 9y^2 - 18xy - 6x - 6y + 3 = 0 \]

### Step 10: Divide by 3 for simplicity
Dividing the entire equation by 3:
\[ 6x^2 + 3y^2 - 6xy - 2x - 2y + 1 = 0 \]

### Final Result
The locus of the centroid G as α varies is given by the equation:
\[ 6x^2 + 3y^2 - 6xy - 2x - 2y + 1 = 0 \]