The value of k so that `x+3y+k=0` touches the circle `x^(2)+y^(2)+6x+2y=0` is

Options

A. `10`
B. `8`
C. `16`
D. `4`

Answer

To find the value of \( k \) such that the line \( x + 3y + k = 0 \) touches the circle given by the equation \( x^2 + y^2 + 6x + 2y = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 + 6x + 2y = 0 \]
We can complete the square for both \( x \) and \( y \).

For \( x \):
\[ x^2 + 6x = (x + 3)^2 - 9 \]

For \( y \):
\[ y^2 + 2y = (y + 1)^2 - 1 \]

Substituting these back into the equation gives:
\[ (x + 3)^2 - 9 + (y + 1)^2 - 1 = 0 \]
Simplifying this, we have:
\[ (x + 3)^2 + (y + 1)^2 = 10 \]

### Step 2: Identify the Center and Radius of the Circle
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we can identify:
- Center \( C(-3, -1) \)
- Radius \( r = \sqrt{10} \)

### Step 3: Find the Length of the Perpendicular from the Center to the Line
The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( x + 3y + k = 0 \), we have:
- \( A = 1 \)
- \( B = 3 \)
- \( C = k \)

Substituting the center \( C(-3, -1) \):
\[ d = \frac{|1(-3) + 3(-1) + k|}{\sqrt{1^2 + 3^2}} = \frac{|-3 - 3 + k|}{\sqrt{10}} = \frac{|k - 6|}{\sqrt{10}} \]

### Step 4: Set the Distance Equal to the Radius
Since the line touches the circle, the distance \( d \) must equal the radius \( r \):
\[ \frac{|k - 6|}{\sqrt{10}} = \sqrt{10} \]

### Step 5: Solve for \( k \)
Multiplying both sides by \( \sqrt{10} \):
\[ |k - 6| = 10 \]
This gives us two cases:
1. \( k - 6 = 10 \) → \( k = 16 \)
2. \( k - 6 = -10 \) → \( k = -4 \)

### Conclusion
The values of \( k \) such that the line touches the circle are \( k = 16 \) and \( k = -4 \). However, since we are looking for the value of \( k \) that allows the line to touch the circle, we can conclude that:
\[ \text{The value of } k \text{ is } 16. \]