Find the value of K if the lines `x+y-5=0` and `2x+ky-8=0` are conjugate with respect to the circle `x^(2)+y^(2)-2x-2y-1=0`

Answer

To find the value of \( K \) such that the lines \( x + y - 5 = 0 \) and \( 2x + ky - 8 = 0 \) are conjugate with respect to the circle given by \( x^2 + y^2 - 2x - 2y - 1 = 0 \), we will follow these steps:

### Step 1: Identify the center and radius of the circle

The standard form of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
From the equation of the circle \( x^2 + y^2 - 2x - 2y - 1 = 0 \), we can rewrite it as:
\[ x^2 - 2x + y^2 - 2y = 1 \]
Completing the square for \( x \) and \( y \):
\[ (x - 1)^2 + (y - 1)^2 = 3 \]
Thus, the center of the circle is \( (1, 1) \) and the radius \( r \) is \( \sqrt{3} \).

### Step 2: Use the condition for conjugate lines

The condition for two lines \( l_1: l_1x + m_1y + n_1 = 0 \) and \( l_2: l_2x + m_2y + n_2 = 0 \) to be conjugate with respect to the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ r^2 (l_1 l_2 + m_1 m_2) = (l_1 g + m_1 f - n_1)(l_2 g + m_2 f - n_2) \]

### Step 3: Identify coefficients from the lines

From the first line \( x + y - 5 = 0 \):
- \( l_1 = 1 \)
- \( m_1 = 1 \)
- \( n_1 = -5 \)

From the second line \( 2x + ky - 8 = 0 \):
- \( l_2 = 2 \)
- \( m_2 = k \)
- \( n_2 = -8 \)

### Step 4: Substitute values into the conjugate condition

We already found that \( r^2 = 3 \) (since \( r = \sqrt{3} \)). Now substituting into the conjugate condition:
\[ 3(1 \cdot 2 + 1 \cdot k) = (1 \cdot (-1) + 1 \cdot (-1) - (-5))(2 \cdot (-1) + k \cdot (-1) - (-8)) \]
This simplifies to:
\[ 3(2 + k) = (-1 - 1 + 5)(-2 - k + 8) \]
\[ 3(2 + k) = (3)(6 - k) \]

### Step 5: Solve for \( k \)

Expanding both sides:
\[ 6 + 3k = 18 - 3k \]
Bringing all terms involving \( k \) to one side:
\[ 3k + 3k = 18 - 6 \]
\[ 6k = 12 \]
\[ k = 2 \]

### Final Answer
The value of \( K \) is \( 2 \).

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