The vertices of a triangle ABC has co -ordinates `(cos theta, sin theta)`
`(sin theta, - cos theta),(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 these

Answer

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

### Step 1: Find the Coordinates of the Centroid
The coordinates of the centroid \( G(h, k) \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are given by:

\[ h = \frac{x_1 + x_2 + x_3}{3}, \quad k = \frac{y_1 + y_2 + y_3}{3} \]

For our triangle:

- \( A(\cos \theta, \sin \theta) \)
- \( B(\sin \theta, -\cos \theta) \)
- \( C(1, 2) \)

Substituting these coordinates into the centroid formula:

\[ h = \frac{\cos \theta + \sin \theta + 1}{3} \]

\[ k = \frac{\sin \theta - \cos \theta + 2}{3} \]

### Step 2: Express \( h \) and \( k \) in terms of \( \theta \)
Now we have:

\[ 3h = \cos \theta + \sin \theta + 1 \]

\[ 3k = \sin \theta - \cos \theta + 2 \]

### Step 3: Rearranging the equations
From the equations above, we can rearrange them:

1. \( 3h - 1 = \cos \theta + \sin \theta \)
2. \( 3k - 2 = \sin \theta - \cos \theta \)

### Step 4: Square both equations
Now we will square both equations and add them:

\[ (3h - 1)^2 = (\cos \theta + \sin \theta)^2 \]

\[ (3k - 2)^2 = (\sin \theta - \cos \theta)^2 \]

### Step 5: Expand both sides
Expanding these equations gives:

1. \( (3h - 1)^2 = \cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta \)
2. \( (3k - 2)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \)

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

1. \( (3h - 1)^2 = 1 + 2\sin \theta \cos \theta \)
2. \( (3k - 2)^2 = 1 - 2\sin \theta \cos \theta \)

### Step 6: Add the two equations
Now, adding these two equations:

\[ (3h - 1)^2 + (3k - 2)^2 = 1 + 2\sin \theta \cos \theta + 1 - 2\sin \theta \cos \theta \]

This simplifies to:

\[ (3h - 1)^2 + (3k - 2)^2 = 2 \]

### Step 7: Rewrite in standard form
Now, let’s rewrite this in standard form:

\[ (3h - 1)^2 + (3k - 2)^2 = 2 \]

### Step 8: Identify the locus
This equation represents a circle with center \( \left(\frac{1}{3}, \frac{2}{3}\right) \) and radius \( \sqrt{2} \).

### Conclusion
Thus, as \( \theta \) varies, the locus of the centroid of triangle ABC is a circle.

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