Locus of the point `(cos theta +sin theta, cos theta - sin theta)` where `theta` is parameter is

Options

A. `x^(2)+y^(2)=1`
B. `x^(2)+y^(2)=4`
C. `x^(2)+y^(2)=2`
D. `y^(2)=4ax`

Answer

To find the locus of the point \((\cos \theta + \sin \theta, \cos \theta - \sin \theta)\) where \(\theta\) is a parameter, we can follow these steps:

### Step 1: Define the Coordinates
Let:
- \(x = \cos \theta + \sin \theta\) (Equation 1)
- \(y = \cos \theta - \sin \theta\) (Equation 2)

### Step 2: Square Both Equations
Now, we will square both equations:
- From Equation 1:
\[ x^2 = (\cos \theta + \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta + 2\cos \theta \sin \theta \]
- From Equation 2:
\[ y^2 = (\cos \theta - \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta - 2\cos \theta \sin \theta \]

### Step 3: Add the Squared Equations
Now, add the two squared equations:
\[ x^2 + y^2 = \left(\cos^2 \theta + \sin^2 \theta + 2\cos \theta \sin \theta\right) + \left(\cos^2 \theta + \sin^2 \theta - 2\cos \theta \sin \theta\right) \]

### Step 4: Simplify the Expression
Combine like terms:
\[ x^2 + y^2 = 2(\cos^2 \theta + \sin^2 \theta) \]
Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ x^2 + y^2 = 2 \cdot 1 = 2 \]

### Conclusion
Thus, the locus of the point is given by:
\[ x^2 + y^2 = 2 \]

### Final Answer
The correct option is \(x^2 + y^2 = 2\).

---