If `3x+2y=3` and `2x+5y=1` are conjugate lines w.r.t the circle `x^(2)+y^(2)=r^(2)` then `r^(2)=`

Options

A. `3/16`
B. `16/3`
C. `4/(sqrt(3))`
D. `(sqrt(3))/4`

Answer

To solve the problem, we need to find the value of \( r^2 \) given that the lines \( 3x + 2y = 3 \) and \( 2x + 5y = 1 \) are conjugate with respect to the circle \( x^2 + y^2 = r^2 \).

### Step-by-Step Solution:

1. **Rewrite the equations of the lines in standard form**:
- The first line \( 3x + 2y = 3 \) can be rewritten as:
\[ 3x + 2y - 3 = 0 \]
Here, \( L_1 = 3 \), \( M_1 = 2 \), and \( N_1 = -3 \).

- The second line \( 2x + 5y = 1 \) can be rewritten as:
\[ 2x + 5y - 1 = 0 \]
Here, \( L_2 = 2 \), \( M_2 = 5 \), and \( N_2 = -1 \).

2. **Identify the coefficients for the circle**:
- The equation of the circle is \( x^2 + y^2 = r^2 \), which can be rewritten as:
\[ x^2 + y^2 + 0x + 0y - r^2 = 0 \]
Thus, \( g = 0 \), \( f = 0 \), and \( c = -r^2 \).

3. **Use the condition for conjugate lines**:
- The condition for the lines to be conjugate with respect to the circle is given by:
\[ r^2 (L_1 L_2 + M_1 M_2) = L_1 g + M_1 f - N_1 (L_2 g + M_2 f - N_2) \]
- Substituting the values we have:
\[ r^2 (3 \cdot 2 + 2 \cdot 5) = 3 \cdot 0 + 2 \cdot 0 - (-3)(2 \cdot 0 + 5 \cdot 0 - (-1)) \]

4. **Calculate the left-hand side**:
- Calculate \( L_1 L_2 + M_1 M_2 \):
\[ 3 \cdot 2 + 2 \cdot 5 = 6 + 10 = 16 \]
- Thus, the left-hand side becomes:
\[ r^2 \cdot 16 \]

5. **Calculate the right-hand side**:
- The right-hand side simplifies to:
\[ 0 + 0 + 3 \cdot 1 = 3 \]

6. **Set the left-hand side equal to the right-hand side**:
- We have:
\[ 16r^2 = 3 \]

7. **Solve for \( r^2 \)**:
- Dividing both sides by 16 gives:
\[ r^2 = \frac{3}{16} \]

### Final Answer:
Thus, the value of \( r^2 \) is:
\[ \boxed{\frac{3}{16}} \]