The locus of the points of intersection of the lines `x cos theta+y sin theta=a ` and `x sin theta-y cos theta=b`, (`theta=` variable) is :

Options

A. `x^(2)+y^(2)=a^(2)+b^(2)`
B. `x^(2)+y^(2)=a^(2)-b^(2)`
C. `x^(2)+y^(2)=2(a^(2)+b^(2))`
D. None of the above

Answer

To find the locus of the points of intersection of the lines given by the equations \( x \cos \theta + y \sin \theta = a \) and \( x \sin \theta - y \cos \theta = b \), we will follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( x \cos \theta + y \sin \theta = a \) (Equation 1)
2. \( x \sin \theta - y \cos \theta = b \) (Equation 2)

### Step 2: Multiply the equations by appropriate trigonometric functions
To eliminate \( x \) and \( y \), we can manipulate these equations. Multiply Equation 1 by \( \sin \theta \):
\[ x \sin \theta \cos \theta + y \sin^2 \theta = a \sin \theta \]
Multiply Equation 2 by \( \cos \theta \):
\[ x \sin \theta \cos \theta - y \cos^2 \theta = b \cos \theta \]

### Step 3: Subtract the modified equations
Now, subtract the second modified equation from the first:
\[ (x \sin \theta \cos \theta + y \sin^2 \theta) - (x \sin \theta \cos \theta - y \cos^2 \theta) = a \sin \theta - b \cos \theta \]
This simplifies to:
\[ y (\sin^2 \theta + \cos^2 \theta) = a \sin \theta - b \cos \theta \]

### Step 4: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ y = a \sin \theta - b \cos \theta \]

### Step 5: Substitute \( y \) back into one of the original equations
Now substitute \( y \) back into Equation 1:
\[ x \cos \theta + (a \sin \theta - b \cos \theta) \sin \theta = a \]
This simplifies to:
\[ x \cos \theta + a \sin^2 \theta - b \sin \theta \cos \theta = a \]
Rearranging gives:
\[ x \cos \theta = a - a \sin^2 \theta + b \sin \theta \cos \theta \]

### Step 6: Solve for \( x \)
Now, factor out \( a \):
\[ x = \frac{a(1 - \sin^2 \theta) + b \sin \theta \cos \theta}{\cos \theta} \]
Using \( 1 - \sin^2 \theta = \cos^2 \theta \):
\[ x = \frac{a \cos^2 \theta + b \sin \theta \cos \theta}{\cos \theta} \]
This simplifies to:
\[ x = a \cos \theta + b \sin \theta \]

### Step 7: Find the locus by eliminating \( \theta \)
We have:
\[ x = a \cos \theta + b \sin \theta \]
\[ y = a \sin \theta - b \cos \theta \]
Now, square both equations:
\[ x^2 = (a \cos \theta + b \sin \theta)^2 \]
\[ y^2 = (a \sin \theta - b \cos \theta)^2 \]

### Step 8: Add the squared equations
Adding these gives:
\[ x^2 + y^2 = (a^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta) \]
This simplifies to:
\[ x^2 + y^2 = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) \]
Using \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ x^2 + y^2 = a^2 + b^2 \]

### Final Answer
The locus of the points of intersection is:
\[ x^2 + y^2 = a^2 + b^2 \]