Let the centre of a circle C be (alpha, beta ) and its radius `r<8`. Let `3x +4y =24` and `3x -4y =32` be two tangents and `4x +3y =1` be a normal to C Then `(alpha -beta +r)` is equal to .

Options

A. `6`
B. `7`
C. `5`
D. `9`

Answer

To solve the problem, we need to find the value of \( \alpha - \beta + r \) given the conditions of the circle, tangents, and normal. Let's break down the solution step by step.

### Step 1: Identify the equations of the tangents
The equations of the tangents are given as:
1. \( 3x + 4y = 24 \)
2. \( 3x - 4y = 32 \)

### Step 2: Use the formula for the distance from a point to a line
The distance \( d \) from the center of the circle \( ( \alpha, \beta ) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|A\alpha + B\beta + C|}{\sqrt{A^2 + B^2}} \]

For the first tangent \( 3x + 4y - 24 = 0 \):
- \( A = 3, B = 4, C = -24 \)
- The distance is:
\[ d_1 = \frac{|3\alpha + 4\beta - 24|}{\sqrt{3^2 + 4^2}} = \frac{|3\alpha + 4\beta - 24|}{5} \]

For the second tangent \( 3x - 4y - 32 = 0 \):
- \( A = 3, B = -4, C = -32 \)
- The distance is:
\[ d_2 = \frac{|3\alpha - 4\beta - 32|}{\sqrt{3^2 + (-4)^2}} = \frac{|3\alpha - 4\beta - 32|}{5} \]

### Step 3: Set the distances equal to the radius \( r \)
Since both lines are tangents to the circle, we have:
\[ \frac{|3\alpha + 4\beta - 24|}{5} = r \]
\[ \frac{|3\alpha - 4\beta - 32|}{5} = r \]

### Step 4: Equate the two expressions
From the above equations, we can set them equal:
\[ |3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32| \]

### Step 5: Solve the absolute value equation
This gives us two cases to consider:

**Case 1:**
\[ 3\alpha + 4\beta - 24 = 3\alpha - 4\beta - 32 \]
This simplifies to:
\[ 8\beta = -8 \implies \beta = -1 \]

**Case 2:**
\[ 3\alpha + 4\beta - 24 = -(3\alpha - 4\beta - 32) \]
This simplifies to:
\[ 3\alpha + 4\beta - 24 = -3\alpha + 4\beta + 32 \]
\[ 6\alpha = 56 \implies \alpha = \frac{28}{3} \]

### Step 6: Find the normal to the circle
The normal to the circle is given by the line \( 4x + 3y = 1 \). The normal at the center \( (\alpha, \beta) \) gives us:
\[ 4\alpha + 3\beta = 1 \]
Substituting \( \beta = -1 \):
\[ 4\alpha + 3(-1) = 1 \implies 4\alpha - 3 = 1 \implies 4\alpha = 4 \implies \alpha = 1 \]

### Step 7: Calculate the radius \( r \)
Using the value of \( \alpha \) and \( \beta \) in the distance formula for the first tangent:
\[ r = \frac{|3(1) + 4(-1) - 24|}{5} = \frac{|3 - 4 - 24|}{5} = \frac{|-25|}{5} = 5 \]

### Step 8: Calculate \( \alpha - \beta + r \)
Now we can find:
\[ \alpha - \beta + r = 1 - (-1) + 5 = 1 + 1 + 5 = 7 \]

### Final Answer
Thus, the value of \( \alpha - \beta + r \) is \( 7 \).

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