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 \( \lambda \) so that the line \( 3x - 4y = \lambda \) touches the circle given by the equation \( x^2 + y^2 - 4x - 8y - 5 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 8y - 5 = 0 \]
We can rewrite this by completing 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 back into the equation gives:
\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 - 5 = 0 \]
Simplifying this, we have:
\[ (x - 2)^2 + (y - 4)^2 - 25 = 0 \]
Thus, the standard form 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 distance from the center of the circle to the line
The line can be rewritten in the standard form \( Ax + By + C = 0 \):
\[ 3x - 4y - \lambda = 0 \]
Here, \( A = 3 \), \( B = -4 \), and \( C = -\lambda \).

The distance \( d \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting the center of the circle \( (2, 4) \):
\[ d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = \frac{| -10 - \lambda |}{5} \]

### Step 3: Set the distance equal to the radius
For the line to touch 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 the absolute value equation
This absolute value equation leads to two cases:
1. \( -10 - \lambda = 25 \)
2. \( -10 - \lambda = -25 \)

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

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

### Conclusion
The values of \( \lambda \) for which the line touches the circle are:
\[ \lambda = 15 \quad \text{and} \quad \lambda = -35 \]