If the line `3x-4y-lambda=0` touches the circle `x^2 + y^2-4x-8y- 5=0` at (a, b) then which of the following is not the possible value of `lambda+a + b`?

Options

A. 20
B. `-28`
C. `-30`
D. none of these

Explanation

To solve the problem, we need to determine the values of \( \lambda + a + b \) given the line \( 3x - 4y - \lambda = 0 \) touches the circle defined by \( x^2 + y^2 - 4x - 8y - 5 = 0 \).

### Step 1: Rewrite the Circle Equation
First, we rewrite the circle equation in standard form. The given equation is:
\[ x^2 + y^2 - 4x - 8y - 5 = 0 \]
We can complete the square for both \( x \) and \( y \):

- For \( x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
- For \( y \):
\[ y^2 - 8y = (y - 4)^2 - 16 \]

Substituting these into the circle equation gives:
\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 - 5 = 0 \]
\[ (x - 2)^2 + (y - 4)^2 - 25 = 0 \]
Thus, the equation of the circle is:
\[ (x - 2)^2 + (y - 4)^2 = 25 \]
This shows that the center of the circle is \( (2, 4) \) and the radius \( r = 5 \).

### Step 2: Find the Perpendicular Distance from the Center to the Line
The line equation is:
\[ 3x - 4y - \lambda = 0 \]
The formula for the perpendicular 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, \( A = 3 \), \( B = -4 \), \( C = -\lambda \), and the center of the circle \( (x_0, y_0) = (2, 4) \).

Calculating the distance:
\[ d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = \frac{|-10 - \lambda|}{5} \]

### Step 3: Set the Distance Equal to the Radius
Since the line is tangent to the circle, the distance \( d \) must equal the radius \( r \):
\[ \frac{|-10 - \lambda|}{5} = 5 \]
Multiplying both sides by 5 gives:
\[ |-10 - \lambda| = 25 \]

### Step 4: Solve for \( \lambda \)
This absolute value equation gives us two cases:
1. \( -10 - \lambda = 25 \)
2. \( -10 - \lambda = -25 \)

For the first case:
\[ -10 - \lambda = 25 \implies \lambda = -35 \]

For the second case:
\[ -10 - \lambda = -25 \implies \lambda = 15 \]

### Step 5: Find Points \( (a, b) \)
Now we substitute these values of \( \lambda \) back into the line equation to find \( a \) and \( b \).

For \( \lambda = 15 \):
\[ 3a - 4b - 15 = 0 \implies 3a - 4b = 15 \quad \text{(1)} \]

For \( \lambda = -35 \):
\[ 3a - 4b + 35 = 0 \implies 3a - 4b = -35 \quad \text{(2)} \]

### Step 6: Find \( \lambda + a + b \)
From equation (1):
\[ 3a - 4b = 15 \implies a = \frac{4b + 15}{3} \]
Substituting this into \( \lambda + a + b \):
\[ \lambda + a + b = 15 + \frac{4b + 15}{3} + b \]
Let’s simplify:
\[ = 15 + \frac{4b + 15 + 3b}{3} = 15 + \frac{7b + 15}{3} \]
Calculating for \( b = 0 \) gives:
\[ = 15 + \frac{15}{3} = 15 + 5 = 20 \]

For \( \lambda = -35 \):
\[ 3a - 4b = -35 \implies a = \frac{4b - 35}{3} \]
Then:
\[ \lambda + a + b = -35 + \frac{4b - 35}{3} + b \]
Simplifying gives:
\[ = -35 + \frac{4b - 35 + 3b}{3} = -35 + \frac{7b - 35}{3} \]
Calculating for \( b = 0 \) gives:
\[ = -35 + \frac{-35}{3} = -35 - \frac{35}{3} = -\frac{105 + 35}{3} = -\frac{140}{3} \approx -30 \]

### Conclusion
The possible values of \( \lambda + a + b \) are \( 20 \) and \( -30 \). The question asks for the value that is **not** possible, which is \( -28 \).

### Final Answer
The value of \( \lambda + a + b \) that is **not** possible is \( -28 \).