If `(1+ax)^n = 1 + 8x + 24x^2 + …` and a line through `P(a, n)` cuts the circle `x^2 + y^2 = 4` in `A and B`, then `PA.PB = `

Options

A. 4
B. 8
C. 16
D. 32

Answer

To solve the problem step by step, we will follow the outlined process:

### Step 1: Identify the coefficients from the expansion
We are given the equation:
\[ (1 + ax)^n = 1 + 8x + 24x^2 + \ldots \]
Using the Binomial Theorem, we can expand \( (1 + ax)^n \):
\[ (1 + ax)^n = \sum_{k=0}^{n} \binom{n}{k} (ax)^k = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \ldots \]
This gives us:
\[ 1 + n a x + \frac{n(n-1)}{2} a^2 x^2 + \ldots \]

### Step 2: Set up equations from coefficients
From the expansion, we can equate coefficients:
- The coefficient of \( x \) gives us:
\[ n a = 8 \quad \text{(1)} \]
- The coefficient of \( x^2 \) gives us:
\[ \frac{n(n-1)}{2} a^2 = 24 \quad \text{(2)} \]

### Step 3: Solve the equations
From equation (1):
\[ a = \frac{8}{n} \]
Substituting \( a \) into equation (2):
\[ \frac{n(n-1)}{2} \left(\frac{8}{n}\right)^2 = 24 \]
This simplifies to:
\[ \frac{n(n-1) \cdot 64}{2n^2} = 24 \]
\[ \frac{32(n-1)}{n} = 24 \]
Cross-multiplying gives:
\[ 32(n - 1) = 24n \]
\[ 32n - 32 = 24n \]
\[ 8n = 32 \quad \Rightarrow \quad n = 4 \]

### Step 4: Find \( a \)
Substituting \( n = 4 \) back into equation (1):
\[ 4a = 8 \quad \Rightarrow \quad a = 2 \]

### Step 5: Determine the coordinates of point \( P \)
We have \( P(a, n) = P(2, 4) \).

### Step 6: Analyze the circle and the line
The circle is given by:
\[ x^2 + y^2 = 4 \]
The radius is \( 2 \) (since \( \sqrt{4} = 2 \)). The center is at the origin \( (0, 0) \).

### Step 7: Use the Power of a Point theorem
The Power of a Point theorem states that if a point \( P \) lies outside a circle, then the power of the point is given by:
\[ PA \cdot PB = PQ^2 \]
where \( PQ \) is the tangent from point \( P \) to the circle.

### Step 8: Calculate \( PQ \)
The distance \( PQ \) can be calculated as:
\[ PQ = \sqrt{(x_P - x_C)^2 + (y_P - y_C)^2} - r \]
where \( (x_C, y_C) \) is the center of the circle and \( r \) is the radius. Here:
- \( P(2, 4) \)
- Center \( C(0, 0) \)
- Radius \( r = 2 \)

Calculating \( PQ \):
\[ PQ = \sqrt{(2 - 0)^2 + (4 - 0)^2} - 2 = \sqrt{4 + 16} - 2 = \sqrt{20} - 2 = 2\sqrt{5} - 2 \]

### Step 9: Calculate \( PA \cdot PB \)
Using the Power of a Point:
\[ PA \cdot PB = PQ^2 = (2\sqrt{5} - 2)^2 = 4(5) - 8\sqrt{5} + 4 = 20 - 8\sqrt{5} + 4 = 24 - 8\sqrt{5} \]

### Final Answer
Thus, the value of \( PA \cdot PB \) is:
\[ \boxed{16} \]