If the circles `x^(2)+y^(2)+2x+2ky+6=0` and `x^(2)+y^(2)+2ky+k=0` intersect orthogonally, then k is

Options

A. `2 or -(3)/(2)`
B. `-2 or -(3)/(2)`
C. `2 or (3)/(2)`
D. `-2 or (3)/(2)`

Answer

To solve the problem, we need to determine the value of \( k \) such that the circles given by the equations \( x^2 + y^2 + 2x + 2ky + 6 = 0 \) and \( x^2 + y^2 + 2ky + k = 0 \) intersect orthogonally.

### Step-by-Step Solution:

1. **Identify the General Form of the Circle:**
The general equation of a circle is given by:
\[ x^2 + y^2 + 2Gx + 2Fy + C = 0 \]
From the equations, we can identify the coefficients for both circles.

2. **Circle 1:**
The first circle is:
\[ x^2 + y^2 + 2x + 2ky + 6 = 0 \]
Here, we have:
- \( G_1 = 1 \)
- \( F_1 = k \)
- \( C_1 = 6 \)

3. **Circle 2:**
The second circle is:
\[ x^2 + y^2 + 2ky + k = 0 \]
Here, we have:
- \( G_2 = 0 \)
- \( F_2 = k \)
- \( C_2 = k \)

4. **Condition for Orthogonality:**
The circles intersect orthogonally if the following condition holds:
\[ G_1G_2 + F_1F_2 = C_1 + C_2 \]
Substituting the values we found:
\[ (1)(0) + (k)(k) = 6 + k \]
This simplifies to:
\[ k^2 = 6 + k \]

5. **Rearranging the Equation:**
Rearranging gives us:
\[ k^2 - k - 6 = 0 \]

6. **Factoring the Quadratic Equation:**
We can factor this quadratic equation:
\[ (k - 3)(k + 2) = 0 \]
Thus, the solutions for \( k \) are:
\[ k = 3 \quad \text{or} \quad k = -2 \]

7. **Final Answer:**
The values of \( k \) that satisfy the condition of orthogonality are \( k = 3 \) and \( k = -2 \).