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 eliminate the parameter \( \theta \) and derive a relationship between \( x \) and \( y \).

### Step-by-Step Solution:

1. **Define the Coordinates:**
\[ x = \cos \theta + \sin \theta \]
\[ y = \sin \theta - \cos \theta \]

2. **Square Both Equations:**
- For \( x \):
\[ x^2 = (\cos \theta + \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ x^2 = 1 + 2\sin \theta \cos \theta \]
We can express \( 2\sin \theta \cos \theta \) as \( \sin 2\theta \):
\[ x^2 = 1 + \sin 2\theta \]

3. **Square the Second Equation:**
- For \( y \):
\[ y^2 = (\sin \theta - \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \]
Again using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ y^2 = 1 - 2\sin \theta \cos \theta \]
Which can also be expressed as:
\[ y^2 = 1 - \sin 2\theta \]

4. **Combine the Two Equations:**
Now we have:
\[ x^2 = 1 + \sin 2\theta \quad \text{(1)} \]
\[ y^2 = 1 - \sin 2\theta \quad \text{(2)} \]
Adding equations (1) and (2):
\[ x^2 + y^2 = (1 + \sin 2\theta) + (1 - \sin 2\theta) \]
The \( \sin 2\theta \) terms cancel out:
\[ x^2 + y^2 = 2 \]

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