If the lines `2x+y+12=0, 4x-3y-10=0` are conjugate w.r.t the circle with centre `(2,(-3)/2)` then r=

Options

A. `(sqrt(29))/2`
B. `sqrt(29)`
C. `5/2`
D. `5`

Answer

To solve the problem, we need to find the radius \( r \) of the circle given that the lines \( 2x + y + 12 = 0 \) and \( 4x - 3y - 10 = 0 \) are conjugate with respect to the circle centered at \( (2, -\frac{3}{2}) \).

### Step-by-Step Solution:

1. **Identify the lines and their coefficients**:
- The first line is \( l_1: 2x + y + 12 = 0 \) with coefficients \( l_1 = 2, m_1 = 1, n_1 = 12 \).
- The second line is \( l_2: 4x - 3y - 10 = 0 \) with coefficients \( l_2 = 4, m_2 = -3, n_2 = -10 \).

2. **Identify the center of the circle**:
- The center of the circle is given as \( (2, -\frac{3}{2}) \).
- From the standard circle equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we have \( g = -2 \) and \( f = \frac{3}{2} \).

3. **Use the conjugate condition**:
- The condition for two lines to be conjugate with respect to a circle is given by:
\[ r^2 (l_1 l_2 + m_1 m_2) = l_1 g + m_1 f - n_1 \cdot (l_2 g + m_2 f - n_2) \]

4. **Calculate \( l_1 l_2 + m_1 m_2 \)**:
- \( l_1 l_2 = 2 \cdot 4 = 8 \)
- \( m_1 m_2 = 1 \cdot (-3) = -3 \)
- Therefore, \( l_1 l_2 + m_1 m_2 = 8 - 3 = 5 \).

5. **Calculate \( l_1 g + m_1 f - n_1 \)**:
- \( l_1 g = 2 \cdot (-2) = -4 \)
- \( m_1 f = 1 \cdot \frac{3}{2} = \frac{3}{2} \)
- So, \( l_1 g + m_1 f - n_1 = -4 + \frac{3}{2} - 12 \).
- Converting \( -4 \) and \( -12 \) to halves: \( -4 = -\frac{8}{2} \) and \( -12 = -\frac{24}{2} \).
- Thus, \( -\frac{8}{2} + \frac{3}{2} - \frac{24}{2} = -\frac{29}{2} \).

6. **Calculate \( l_2 g + m_2 f - n_2 \)**:
- \( l_2 g = 4 \cdot (-2) = -8 \)
- \( m_2 f = -3 \cdot \frac{3}{2} = -\frac{9}{2} \)
- So, \( l_2 g + m_2 f - n_2 = -8 - \frac{9}{2} + 10 \).
- Converting \( -8 \) to halves: \( -8 = -\frac{16}{2} \).
- Thus, \( -\frac{16}{2} - \frac{9}{2} + \frac{20}{2} = -\frac{5}{2} \).

7. **Substituting into the conjugate condition**:
- We have:
\[ r^2 \cdot 5 = -\frac{29}{2} \cdot -\frac{5}{2} \]
- This simplifies to:
\[ r^2 \cdot 5 = \frac{145}{4} \]
- Therefore:
\[ r^2 = \frac{145}{20} = \frac{29}{4} \]

8. **Finding the radius \( r \)**:
- Taking the square root:
\[ r = \sqrt{\frac{29}{4}} = \frac{\sqrt{29}}{2} \]

### Final Answer:
The radius \( r \) of the circle is \( \frac{\sqrt{29}}{2} \).