If `2x+3y=1` and `3x+4y=k` are conjugate lines w.r. the circle `x^(2)+y^(2)=4` then k=

Options

A. 36
B. 72
C. 24
D. 144

Answer

To solve the problem, we need to find the value of \( k \) such that the lines \( 2x + 3y = 1 \) and \( 3x + 4y = k \) are conjugate with respect to the circle \( x^2 + y^2 = 4 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the first line**:
The first line is given by the equation \( 2x + 3y = 1 \). We can rewrite it in the standard form:
\[ 2x + 3y - 1 = 0 \]
From this, we can identify:
- \( l_1 = 2 \)
- \( m_1 = 3 \)
- \( n_1 = -1 \)

2. **Identify the coefficients of the second line**:
The second line is given by the equation \( 3x + 4y = k \). Rewriting it in standard form gives:
\[ 3x + 4y - k = 0 \]
From this, we can identify:
- \( l_2 = 3 \)
- \( m_2 = 4 \)
- \( n_2 = -k \)

3. **Identify the radius of the circle**:
The equation of the circle is given by \( x^2 + y^2 = 4 \). This can be compared to the general form \( x^2 + y^2 = r^2 \), where \( r \) is the radius. From this, we find:
\[ r = 2 \]

4. **Use the condition for conjugate lines**:
The condition for two lines to be conjugate with respect to a circle is given by:
\[ r^2 l_1 l_2 + m_1 m_2 = n_1 n_2 \]
Substituting the values we have:
- \( r^2 = 2^2 = 4 \)
- \( l_1 = 2 \)
- \( l_2 = 3 \)
- \( m_1 = 3 \)
- \( m_2 = 4 \)
- \( n_1 = -1 \)
- \( n_2 = -k \)

Plugging these into the equation:
\[ 4 \cdot (2 \cdot 3) + (3 \cdot 4) = (-1)(-k) \]
Simplifying the left side:
\[ 4 \cdot 6 + 12 = k \]
\[ 24 + 12 = k \]
\[ k = 36 \]

5. **Final Answer**:
Thus, the value of \( k \) is:
\[ k = 36 \]