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

Answer

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

### Step 1: Rewrite the circle equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 4x - 8y - 5 = 0 \]
We can complete the square for \( x \) 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:
\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 - 5 = 0 \]
\[ (x - 2)^2 + (y - 4)^2 - 25 = 0 \]
Thus, the equation of the circle is:
\[ (x - 2)^2 + (y - 4)^2 = 25 \]
This indicates that the center of the circle is \( (2, 4) \) and the radius \( r = 5 \).

### Step 2: Find the distance from the center to the line
The distance \( d \) from the center of the circle \( (2, 4) \) to the line \( 3x - 4y - \lambda = 0 \) is given by the formula:
\[ d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - \lambda|}{5} = \frac{| -10 - \lambda|}{5} \]
Since the line is tangent to the circle, this distance must equal the radius:
\[ \frac{| -10 - \lambda|}{5} = 5 \]
Multiplying both sides by 5:
\[ | -10 - \lambda| = 25 \]

### Step 3: Solve for \( \lambda \)
This absolute value equation gives us two cases:
1. \( -10 - \lambda = 25 \)
\[ \lambda = -35 \]
2. \( -10 - \lambda = -25 \)
\[ \lambda = 15 \]

### Step 4: Find \( a \) and \( b \)
The point of tangency \( (a, b) \) lies on the line \( 3x - 4y - \lambda = 0 \). We can express \( a \) in terms of \( b \):
\[ 3a - 4b - \lambda = 0 \implies 3a = 4b + \lambda \implies a = \frac{4b + \lambda}{3} \]

### Step 5: Calculate \( \lambda + a + b \)
Now we calculate \( \lambda + a + b \) for both values of \( \lambda \):

1. For \( \lambda = 15 \):
\[ a = \frac{4b + 15}{3} \]
Substituting \( b = 0 \) (as derived from the distance condition):
\[ a = \frac{4(0) + 15}{3} = 5 \]
Thus,
\[ \lambda + a + b = 15 + 5 + 0 = 20 \]

2. For \( \lambda = -35 \):
\[ a = \frac{4b - 35}{3} \]
Again substituting \( b = 0 \):
\[ a = \frac{4(0) - 35}{3} = -\frac{35}{3} \]
Thus,
\[ \lambda + a + b = -35 - \frac{35}{3} + 0 = -\frac{105}{3} - \frac{35}{3} = -\frac{140}{3} = -30 \]

### Step 6: Identify the impossible value
The possible values of \( \lambda + a + b \) are \( 20 \) and \( -30 \). The question asks for which of the following is not a possible value for \( \lambda + a + b \). If the options include values like \( -28 \), it is not possible since the only values we derived are \( 20 \) and \( -30 \).

### Conclusion
The value that is not possible for \( \lambda + a + b \) is \( -28 \).