The vertices of a `DeltaABC` has coordinates `(costheta,sintheta),(sintheta,-costheta)` and (1,2). As `theta` varies the locus of centroid of the triangle is the circle

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 the above

Answer

To find the locus of the centroid of triangle ABC with vertices at \((\cos \theta, \sin \theta)\), \((\sin \theta, -\cos \theta)\), and \((1, 2)\) as \(\theta\) varies, we will follow these steps:

### Step 1: Find the Centroid of Triangle ABC
The centroid \(G\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) 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, the vertices are:
- \(A(\cos \theta, \sin \theta)\)
- \(B(\sin \theta, -\cos \theta)\)
- \(C(1, 2)\)

Thus, the coordinates of the centroid \(G\) are:
\[ G\left(\frac{\cos \theta + \sin \theta + 1}{3}, \frac{\sin \theta - \cos \theta + 2}{3}\right) \]

### Step 2: Set the Centroid Coordinates
Let the coordinates of the centroid be \((h, k)\):
\[ h = \frac{\cos \theta + \sin \theta + 1}{3} \]
\[ k = \frac{\sin \theta - \cos \theta + 2}{3} \]

### Step 3: Rearranging the Equations
From the equations for \(h\) and \(k\), we can rearrange them:
1. From \(h\):
\[ \cos \theta + \sin \theta = 3h - 1 \]
2. From \(k\):
\[ \sin \theta - \cos \theta = 3k - 2 \]

### Step 4: Square and Add the Equations
Now we will square both equations and add them:
\[ (\cos \theta + \sin \theta)^2 + (\sin \theta - \cos \theta)^2 = (3h - 1)^2 + (3k - 2)^2 \]

Expanding both sides:
- Left Side:
\[ \cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta + \sin^2 \theta - 2\cos \theta \sin \theta + \cos^2 \theta = 1 + 1 = 2 \]
- Right Side:
\[ (3h - 1)^2 + (3k - 2)^2 = 9h^2 - 6h + 1 + 9k^2 - 12k + 4 = 9h^2 + 9k^2 - 6h - 12k + 5 \]

### Step 5: Set the Equation
Equating both sides gives:
\[ 9h^2 + 9k^2 - 6h - 12k + 5 = 2 \]
This simplifies to:
\[ 9h^2 + 9k^2 - 6h - 12k + 3 = 0 \]

### Step 6: Replace \(h\) and \(k\) with \(x\) and \(y\)
Let \(h = x\) and \(k = y\):
\[ 9x^2 + 9y^2 - 6x - 12y + 3 = 0 \]

### Step 7: Simplify the Equation
Dividing the entire equation by 3 gives:
\[ 3x^2 + 3y^2 - 2x - 4y + 1 = 0 \]

### Step 8: Rearranging to Standard Form
Rearranging gives us:
\[ 3x^2 + 3y^2 - 2x - 4y + 1 = 0 \]

This represents a circle in the coordinate plane.

### Final Result
The locus of the centroid of triangle ABC as \(\theta\) varies is given by the equation:
\[ 3x^2 + 3y^2 - 2x - 4y + 1 = 0 \]

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