If `x^(2)+y^(2)+6x+2ky+25=0` to touch y-axis then k=

Options

A. `+-20`
B. `-1,-5`
C. `+-5`
D. `4`

Answer

To solve the problem, we need to determine the value of \( k \) such that the circle defined by the equation \( x^2 + y^2 + 6x + 2ky + 25 = 0 \) touches the y-axis.

### Step-by-Step Solution:

1. **Identify the Circle's Equation**:
The given equation of the circle is:
\[ x^2 + y^2 + 6x + 2ky + 25 = 0 \]

2. **Rearranging the Equation**:
To find the center and radius of the circle, we can rearrange the equation. First, we can group the \( x \) terms:
\[ (x^2 + 6x) + (y^2 + 2ky) + 25 = 0 \]

3. **Completing the Square for \( x \)**:
For the \( x \) terms, we complete the square:
\[ x^2 + 6x = (x + 3)^2 - 9 \]
Thus, the equation becomes:
\[ (x + 3)^2 - 9 + (y^2 + 2ky) + 25 = 0 \]
Simplifying this gives:
\[ (x + 3)^2 + (y^2 + 2ky) + 16 = 0 \]

4. **Completing the Square for \( y \)**:
Now, we complete the square for the \( y \) terms:
\[ y^2 + 2ky = (y + k)^2 - k^2 \]
Substituting this back into the equation gives:
\[ (x + 3)^2 + ((y + k)^2 - k^2) + 16 = 0 \]
This simplifies to:
\[ (x + 3)^2 + (y + k)^2 + 16 - k^2 = 0 \]

5. **Setting the Equation to Standard Form**:
Rearranging gives:
\[ (x + 3)^2 + (y + k)^2 = k^2 - 16 \]
This represents a circle centered at \( (-3, -k) \) with radius \( \sqrt{k^2 - 16} \).

6. **Condition for Touching the y-axis**:
For the circle to touch the y-axis, the distance from the center of the circle to the y-axis must equal the radius. The distance from the center \( (-3, -k) \) to the y-axis (which is at \( x = 0 \)) is \( 3 \). Therefore, we set up the equation:
\[ \sqrt{k^2 - 16} = 3 \]

7. **Squaring Both Sides**:
Squaring both sides gives:
\[ k^2 - 16 = 9 \]

8. **Solving for \( k \)**:
Adding 16 to both sides:
\[ k^2 = 25 \]
Taking the square root of both sides results in:
\[ k = \pm 5 \]

### Final Answer:
Thus, the values of \( k \) are:
\[ k = 5 \quad \text{or} \quad k = -5 \]