If the coordinates of a variable point be `(cos theta + sin theta, sin theta - cos theta)`, where `theta` is the parameter, then the locus of P is

Options

A. `x^(2)-y^(2)=4`
B. `x^(2)+y^(2)=2`
C. xy = 3
D. `x^(2)+2y^(2)=3`

Answer

To find the locus of the point \( P \) with coordinates \( (x, y) = (\cos \theta + \sin \theta, \sin \theta - \cos \theta) \), we will express \( x \) and \( y \) in terms of \( \theta \) and then eliminate \( \theta \) to find the relationship between \( x \) and \( y \).

### Step 1: Express \( x \) and \( y \)
Given:
\[ x = \cos \theta + \sin \theta \]
\[ y = \sin \theta - \cos \theta \]

### Step 2: Square both equations
Now, we will square both \( x \) and \( y \):
\[ x^2 = (\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ x^2 = 1 + 2\cos \theta \sin \theta \]
We know that \( 2\cos \theta \sin \theta = \sin 2\theta \), so:
\[ x^2 = 1 + \sin 2\theta \]

### Step 3: Square the second equation
Now, square \( y \):
\[ y^2 = (\sin \theta - \cos \theta)^2 = \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta \]
Again using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ y^2 = 1 - 2\sin \theta \cos \theta \]
And since \( 2\sin \theta \cos \theta = \sin 2\theta \):
\[ y^2 = 1 - \sin 2\theta \]

### Step 4: Add the equations
Now, we add \( x^2 \) and \( y^2 \):
\[ x^2 + y^2 = (1 + \sin 2\theta) + (1 - \sin 2\theta) \]
This simplifies to:
\[ x^2 + y^2 = 2 \]

### Conclusion
The locus of the point \( P \) is given by the equation:
\[ x^2 + y^2 = 2 \]
This represents a circle centered at the origin with a radius of \( \sqrt{2} \).