The sum of all the values of `lambda` for which the set `{(x, y):x^(2)+y^(2)-6x+4y=12}nn{(x, y): 4x+3y= lambda}` contains exactly one element is

Options

A. 31
B. `-31`
C. 12
D. `-19`

Answer

To solve the problem, we need to find the sum of all values of \( \lambda \) for which the intersection of the sets defined by the circle and the line contains exactly one point. This implies that the line is tangent to the circle.

### Step 1: Identify the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 6x + 4y = 12 \]
We can rewrite this in standard form by completing the square.

### Step 2: Complete the Square
1. For \( x^2 - 6x \), we complete the square:
\[ x^2 - 6x = (x - 3)^2 - 9 \]
2. For \( y^2 + 4y \), we complete the square:
\[ y^2 + 4y = (y + 2)^2 - 4 \]
3. Substitute back into the circle equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 = 12 \]
Simplifying gives:
\[ (x - 3)^2 + (y + 2)^2 = 25 \]

### Step 3: Identify the Center and Radius
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \):
- Center \( (h, k) = (3, -2) \)
- Radius \( r = 5 \)

### Step 4: Write the Line Equation
The line is given by:
\[ 4x + 3y = \lambda \]
We can rewrite this in standard form:
\[ 4x + 3y - \lambda = 0 \]

### Step 5: Calculate the Distance 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:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 4 \), \( B = 3 \), \( C = -\lambda \), and the center \( (x_0, y_0) = (3, -2) \).

Calculating the distance:
\[ d = \frac{|4(3) + 3(-2) - \lambda|}{\sqrt{4^2 + 3^2}} = \frac{|12 - 6 - \lambda|}{\sqrt{16 + 9}} = \frac{|6 - \lambda|}{5} \]

### Step 6: Set the Distance Equal to the Radius
For the line to be tangent to the circle, the distance must equal the radius:
\[ \frac{|6 - \lambda|}{5} = 5 \]
Multiplying both sides by 5 gives:
\[ |6 - \lambda| = 25 \]

### Step 7: Solve the Absolute Value Equation
This gives us two cases:
1. \( 6 - \lambda = 25 \)
\[ \lambda = 6 - 25 = -19 \]
2. \( 6 - \lambda = -25 \)
\[ \lambda = 6 + 25 = 31 \]

### Step 8: Sum the Values of \( \lambda \)
Now, we sum the two values:
\[ -19 + 31 = 12 \]

### Final Answer
The sum of all values of \( \lambda \) for which the intersection contains exactly one element is:
\[ \boxed{12} \]