The locus of the point of intersection of the lines `x sin theta+ (1 - cos theta)y = a sin theta` and `x sin theta - (1+ cos theta)y + a sin theta= 0` is

Options

A. `x^2 - y^2 = a^2`
B. `x^2 + y^2 = a^2`
C. `y^2 = ax`
D. None of these

Answer

To find the locus of the point of intersection of the lines given by the equations:

1. \( x \sin \theta + (1 - \cos \theta)y = a \sin \theta \) (Equation 1)
2. \( x \sin \theta - (1 + \cos \theta)y + a \sin \theta = 0 \) (Equation 2)

we will follow these steps:

### Step 1: Rewrite the equations
We can rewrite the equations in a more manageable form. For Equation 1, we have:
\[ x \sin \theta + (1 - \cos \theta)y = a \sin \theta \]
For Equation 2, we can rearrange it as:
\[ x \sin \theta - (1 + \cos \theta)y = -a \sin \theta \]

### Step 2: Solve the equations simultaneously
We can solve these two equations simultaneously to find the intersection point. We can add both equations to eliminate \(y\):
\[ (x \sin \theta + (1 - \cos \theta)y) + (x \sin \theta - (1 + \cos \theta)y) = a \sin \theta - a \sin \theta \]
This simplifies to:
\[ 2x \sin \theta = (1 - \cos \theta - (1 + \cos \theta))y \]
This gives:
\[ 2x \sin \theta = -2y \cos \theta \]
Dividing both sides by 2, we have:
\[ x \sin \theta = -y \cos \theta \]
From this, we can express \(y\) in terms of \(x\):
\[ y = -x \frac{\sin \theta}{\cos \theta} = -x \tan \theta \]

### Step 3: Substitute \(y\) back into one of the original equations
Now, we substitute \(y = -x \tan \theta\) back into Equation 1:
\[ x \sin \theta + (1 - \cos \theta)(-x \tan \theta) = a \sin \theta \]
This simplifies to:
\[ x \sin \theta - x (1 - \cos \theta) \frac{\sin \theta}{\cos \theta} = a \sin \theta \]
Factoring out \(x\):
\[ x \left( \sin \theta - (1 - \cos \theta) \frac{\sin \theta}{\cos \theta} \right) = a \sin \theta \]

### Step 4: Simplify the expression
Now we simplify the term in the parentheses:
\[ \sin \theta - \frac{(1 - \cos \theta) \sin \theta}{\cos \theta} = \sin \theta \left( 1 - \frac{(1 - \cos \theta)}{\cos \theta} \right) \]
This becomes:
\[ \sin \theta \left( \frac{\cos \theta - (1 - \cos \theta)}{\cos \theta} \right) = \sin \theta \left( \frac{2\cos \theta - 1}{\cos \theta} \right) \]

### Step 5: Solve for \(x\)
Now we can express \(x\):
\[ x \cdot \sin \theta \cdot \frac{2\cos \theta - 1}{\cos \theta} = a \sin \theta \]
Dividing both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):
\[ x \cdot \frac{2\cos \theta - 1}{\cos \theta} = a \]
Thus:
\[ x = a \cdot \frac{\cos \theta}{2\cos \theta - 1} \]

### Step 6: Find \(y\) in terms of \(x\)
Now substituting \(x\) back into \(y = -x \tan \theta\):
\[ y = -\left( a \cdot \frac{\cos \theta}{2\cos \theta - 1} \right) \cdot \tan \theta \]
This gives:
\[ y = -a \cdot \frac{\sin \theta}{2\cos \theta - 1} \]

### Step 7: Locus equation
Now we have \(x\) and \(y\) in terms of \(\theta\):
\[ x = a \cdot \frac{\cos \theta}{2\cos \theta - 1}, \quad y = -a \cdot \frac{\sin \theta}{2\cos \theta - 1} \]
To find the locus, we eliminate \(\theta\). We can square both equations and add them:
\[ \frac{x^2}{a^2} + \frac{y^2}{a^2} = \frac{\cos^2 \theta + \sin^2 \theta}{(2\cos \theta - 1)^2} \]
Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[ \frac{x^2 + y^2}{a^2} = \frac{1}{(2\cos \theta - 1)^2} \]

### Final Locus Equation
Thus, the locus of the point of intersection is:
\[ x^2 + y^2 = a^2 \]
This represents a circle with radius \(a\).