If the points (k,1) (2,-3) are conjugate w.r.t. `x^(2)+y^(2)+4x-6y-12=0` then k

Options

A. 4
B. 5
C. `5/4`
D. `4/5`

Answer

To find the value of \( k \) such that the points \( (k, 1) \) and \( (2, -3) \) are conjugate with respect to the circle given by the equation \( x^2 + y^2 + 4x - 6y - 12 = 0 \), we will follow these steps:

### Step 1: Rewrite the circle equation
The given equation of the circle is:
\[ x^2 + y^2 + 4x - 6y - 12 = 0 \]
We can rewrite it in the standard form \( (x + g)^2 + (y + f)^2 = r^2 \).

### Step 2: Identify \( g \), \( f \), and \( c \)
From the equation \( x^2 + y^2 + 4x - 6y - 12 = 0 \), we can identify:
- \( g = 2 \) (since \( 2g = 4 \))
- \( f = -3 \) (since \( 2f = -6 \))
- \( c = -12 \)

### Step 3: Use the conjugate points condition
The condition for two points \( (x_1, y_1) \) and \( (x_2, y_2) \) to be conjugate with respect to the circle is given by:
\[ x_1 x_2 + y_1 y_2 + g(x_1 + x_2) + f(y_1 + y_2) + c = 0 \]
Here, we will substitute \( (x_1, y_1) = (k, 1) \) and \( (x_2, y_2) = (2, -3) \).

### Step 4: Substitute the values into the equation
Substituting the values into the conjugate condition:
\[ k \cdot 2 + 1 \cdot (-3) + 2(k + 2) - 3(1 - 3) - 12 = 0 \]
This simplifies to:
\[ 2k - 3 + 2k + 4 + 6 - 12 = 0 \]

### Step 5: Combine like terms
Combining like terms gives:
\[ 4k - 5 = 0 \]

### Step 6: Solve for \( k \)
Now, we solve for \( k \):
\[ 4k = 5 \implies k = \frac{5}{4} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\frac{5}{4}} \]