Find the value of `lambda` so that the line 3x-4y=`lambda` may touch the circle `x^2+y^2-4x-8y-5=0`

Answer

To find the value of \( \lambda \) such that the line \( 3x - 4y = \lambda \) touches the circle given by the equation \( x^2 + y^2 - 4x - 8y - 5 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the circle equation in standard form by completing the square.

The given equation is:
\[ x^2 + y^2 - 4x - 8y - 5 = 0 \]

Group the \( x \) and \( y \) terms:
\[ (x^2 - 4x) + (y^2 - 8y) = 5 \]

Complete the square for \( x \):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

Complete the square for \( y \):
\[ y^2 - 8y = (y - 4)^2 - 16 \]

Substituting back, we have:
\[ ((x - 2)^2 - 4) + ((y - 4)^2 - 16) = 5 \]
\[ (x - 2)^2 + (y - 4)^2 - 20 = 5 \]
\[ (x - 2)^2 + (y - 4)^2 = 25 \]

### Step 2: Identify the Center and Radius of the Circle
From the equation \( (x - 2)^2 + (y - 4)^2 = 25 \), we can identify:
- Center \( C(2, 4) \)
- Radius \( r = \sqrt{25} = 5 \)

### Step 3: Use the Distance Formula
The distance \( d \) from the center of the circle \( C(2, 4) \) to the line \( 3x - 4y = \lambda \) must equal the radius for the line to be tangent to the circle.

The distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

In our case, the line can be rewritten as:
\[ 3x - 4y - \lambda = 0 \]
Here, \( A = 3 \), \( B = -4 \), and \( C = -\lambda \).

Substituting \( (x_1, y_1) = (2, 4) \):
\[ d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = \frac{|-10 - \lambda|}{5} \]

### Step 4: Set the Distance Equal to the Radius
Since the distance must equal the radius:
\[ \frac{|-10 - \lambda|}{5} = 5 \]

Multiplying both sides by 5:
\[ |-10 - \lambda| = 25 \]

### Step 5: Solve the Absolute Value Equation
This gives us two cases to consider:
1. \( -10 - \lambda = 25 \)
2. \( -10 - \lambda = -25 \)

**Case 1:**
\[ -10 - \lambda = 25 \implies -\lambda = 35 \implies \lambda = -35 \]

**Case 2:**
\[ -10 - \lambda = -25 \implies -\lambda = -15 \implies \lambda = 15 \]

### Final Result
The values of \( \lambda \) such that the line touches the circle are:
\[ \lambda = -35 \quad \text{or} \quad \lambda = 15 \]