Find the value of k if `kx+ 3y - 1 = 0,`
`2x + y + 5 = 0` are conjugate lines with
respect to the circle
`x^(2) + y^(2) - 2x - 4y - 4 =0`.

Answer

To find the value of \( k \) such that the lines \( kx + 3y - 1 = 0 \) and \( 2x + y + 5 = 0 \) are conjugate lines with respect to the circle given by the equation \( x^2 + y^2 - 2x - 4y - 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the circle equation in standard form. The given circle equation is:
\[ x^2 + y^2 - 2x - 4y - 4 = 0 \]
We can complete the square for both \( x \) and \( y \).

For \( x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

For \( y \):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these into the circle equation:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 - 4 = 0 \]
\[ (x - 1)^2 + (y - 2)^2 - 9 = 0 \]
Thus, the standard form of the circle is:
\[ (x - 1)^2 + (y - 2)^2 = 3^2 \]
This shows that the circle has center \( (1, 2) \) and radius \( 3 \).

### Step 2: Identify the Lines
The equations of the lines are:
1. \( l_1: kx + 3y - 1 = 0 \)
2. \( l_2: 2x + y + 5 = 0 \)

We can express these lines in the form \( Ax + By + C = 0 \):
- For \( l_1 \): \( A_1 = k, B_1 = 3, C_1 = -1 \)
- For \( l_2 \): \( A_2 = 2, B_2 = 1, C_2 = 5 \)

### Step 3: Use the Conjugate Lines Condition
For the lines to be conjugate with respect to the circle, the following condition must hold:
\[ r^2 (A_1 A_2 + B_1 B_2) = C_1 C_2 \]
Where \( r \) is the radius of the circle. Here, \( r = 3 \), so \( r^2 = 9 \).

Substituting the values:
\[ 9 (k \cdot 2 + 3 \cdot 1) = (-1)(5) \]
This simplifies to:
\[ 9(2k + 3) = -5 \]

### Step 4: Solve for \( k \)
Now we solve the equation:
\[ 9(2k + 3) = -5 \]
Dividing both sides by 9:
\[ 2k + 3 = -\frac{5}{9} \]
Subtracting 3 from both sides:
\[ 2k = -\frac{5}{9} - 3 \]
Converting 3 to a fraction:
\[ 3 = \frac{27}{9} \]
So:
\[ 2k = -\frac{5}{9} - \frac{27}{9} = -\frac{32}{9} \]
Dividing by 2:
\[ k = -\frac{32}{18} = -\frac{16}{9} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{-\frac{16}{9}} \]