If (1,1), (k,2) are conjugate points with respect to the circle `x^(2)+y^(2)+8x+2y+3=0`, then k=?

Options

A. `-12`
B. `(-12)/7`
C. `(-12)/5`
D. `-4`

Answer

To find the value of \( k \) such that the points \( (1, 1) \) and \( (k, 2) \) are conjugate points with respect to the circle given by the equation \( x^2 + y^2 + 8x + 2y + 3 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 + 8x + 2y + 3 = 0 \]
We can compare this with the standard form of a circle equation:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From this, we identify:
- \( g = 4 \) (since \( 2g = 8 \))
- \( f = 1 \) (since \( 2f = 2 \))
- \( c = 3 \)

### Step 2: Use the Conjugate Points Condition
For points \( (x_1, y_1) \) and \( (x_2, y_2) \) to be conjugate points with respect to the circle, the following condition must hold:
\[ x_1 x_2 + y_1 y_2 + g(x_1 + x_2) + f(y_1 + y_2) + c = 0 \]
Here, \( (x_1, y_1) = (1, 1) \) and \( (x_2, y_2) = (k, 2) \).

### Step 3: Substitute the Values into the Condition
Substituting the values into the conjugate points condition:
\[ 1 \cdot k + 1 \cdot 2 + 4(1 + k) + 1(1 + 2) + 3 = 0 \]
This simplifies to:
\[ k + 2 + 4(1 + k) + 3 + 3 = 0 \]

### Step 4: Simplify the Equation
Now, simplifying the left-hand side:
\[ k + 2 + 4 + 4k + 3 + 3 = 0 \]
Combine like terms:
\[ 5k + 12 = 0 \]

### Step 5: Solve for \( k \)
Now, isolate \( k \):
\[ 5k = -12 \\ k = -\frac{12}{5} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{-\frac{12}{5}} \]