P and Q are any two points on the circle `x^2+y^2= 4` such that PQ is a diameter. If `alpha` and `beta` are the lengths of perpendiculars from `P` and `Q` on `x + y = 1` then the maximum value of `alphabeta` is

Options

A. `(1)/(2)`
B. `(7)/(2)`
C. 1
D. 2

Explanation

To solve the problem, we need to find the maximum value of the product of the lengths of the perpendiculars from points \( P \) and \( Q \) on the line \( x + y = 1 \), where \( P \) and \( Q \) are points on the circle defined by the equation \( x^2 + y^2 = 4 \) and \( PQ \) is a diameter of the circle.

### Step-by-Step Solution:

1. **Identify Points on the Circle**:
The circle \( x^2 + y^2 = 4 \) has a center at \( (0, 0) \) and a radius of \( 2 \). We can express points \( P \) and \( Q \) on the circle in terms of an angle \( \theta \):
\[ P = (2 \cos \theta, 2 \sin \theta) \]
\[ Q = (-2 \cos \theta, -2 \sin \theta) \]
This ensures that \( PQ \) is a diameter.

2. **Calculate the Lengths of Perpendiculars**:
The formula for the length of the perpendicular from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( x + y = 1 \), we can rewrite it as \( 1x + 1y - 1 = 0 \) (where \( A = 1, B = 1, C = -1 \)).

- For point \( P \):
\[ \alpha = \frac{|1(2 \cos \theta) + 1(2 \sin \theta) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2 \cos \theta + 2 \sin \theta - 1|}{\sqrt{2}} = \frac{|2(\cos \theta + \sin \theta) - 1|}{\sqrt{2}} \]

- For point \( Q \):
\[ \beta = \frac{|1(-2 \cos \theta) + 1(-2 \sin \theta) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2 \cos \theta - 2 \sin \theta - 1|}{\sqrt{2}} = \frac{|-(2(\cos \theta + \sin \theta) + 1)|}{\sqrt{2}} = \frac{2(\cos \theta + \sin \theta) + 1}{\sqrt{2}} \]

3. **Find the Product \( \alpha \beta \)**:
Now we need to calculate \( \alpha \beta \):
\[ \alpha \beta = \left(\frac{|2(\cos \theta + \sin \theta) - 1|}{\sqrt{2}}\right) \left(\frac{2(\cos \theta + \sin \theta) + 1}{\sqrt{2}}\right) \]
\[ = \frac{|(2(\cos \theta + \sin \theta) - 1)(2(\cos \theta + \sin \theta) + 1)|}{2} \]
\[ = \frac{|4(\cos \theta + \sin \theta)^2 - 1|}{2} \]

4. **Maximize \( \alpha \beta \)**:
Let \( z = \cos \theta + \sin \theta \). We know that:
\[ z^2 \leq 2 \quad \text{(since } \cos^2 \theta + \sin^2 \theta = 1\text{)} \]
Therefore, \( -\sqrt{2} \leq z \leq \sqrt{2} \).

The expression for \( \alpha \beta \) becomes:
\[ \alpha \beta = \frac{|4z^2 - 1|}{2} \]
To maximize \( |4z^2 - 1| \), we can evaluate it at the endpoints:
- If \( z^2 = 2 \):
\[ \alpha \beta = \frac{|4(2) - 1|}{2} = \frac{7}{2} \]
- If \( z^2 = 0 \):
\[ \alpha \beta = \frac{|4(0) - 1|}{2} = \frac{1}{2} \]

5. **Conclusion**:
The maximum value of \( \alpha \beta \) occurs when \( z^2 = 2 \), giving us:
\[ \text{Maximum value of } \alpha \beta = \frac{7}{2} \]

### Final Answer:
The maximum value of \( \alpha \beta \) is \( \frac{7}{2} \).