If `x^(2)+y^(2)-4x-6y+k=0` touches x-axis then k=

Options

A. `+-20`
B. `-1,-5`
C. `+-2`
D. `4`

Answer

To find the value of \( k \) such that the equation \( x^2 + y^2 - 4x - 6y + k = 0 \) represents a circle that touches the x-axis, we can follow these steps:

### Step 1: Rewrite the equation in standard form
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
We need to rewrite the given equation in a similar form. The given equation is:
\[ x^2 + y^2 - 4x - 6y + k = 0 \]
We can rearrange it as:
\[ x^2 - 4x + y^2 - 6y + k = 0 \]

### Step 2: Complete the square for \( x \) and \( y \)
To complete the square for \( x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
To complete the square for \( y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]
Now substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + k = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y - 3)^2 + k - 13 = 0 \]
Thus, we have:
\[ (x - 2)^2 + (y - 3)^2 = 13 - k \]

### Step 3: Identify the center and radius
From the equation \( (x - 2)^2 + (y - 3)^2 = 13 - k \), we can identify:
- The center of the circle is \( (2, 3) \).
- The radius \( r \) is given by \( r = \sqrt{13 - k} \).

### Step 4: Condition for the circle to touch the x-axis
For the circle to touch the x-axis, the distance from the center to the x-axis must equal the radius. The distance from the center \( (2, 3) \) to the x-axis is \( 3 \). Therefore, we set:
\[ r = 3 \]
This gives us the equation:
\[ \sqrt{13 - k} = 3 \]

### Step 5: Solve for \( k \)
Squaring both sides, we get:
\[ 13 - k = 9 \]
Solving for \( k \):
\[ k = 13 - 9 = 4 \]

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