PQ is a diameter of circle `x^2+y^2=4` . If perpendicular distances of P and Q from line `x+y=2` are `alpha` and `beta` respectively then maximum value of `alpha beta` is

Answer

To solve the problem step by step, we will find the maximum value of the product of the perpendicular distances from points P and Q on the circle to the line \(x + y = 2\).

### Step 1: Identify the Circle and Points P and Q
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This is a circle centered at the origin \((0, 0)\) with a radius of \(2\). The points \(P\) and \(Q\) are the endpoints of the diameter of the circle. We can express these points in parametric form:
\[ P = (2 \cos \theta, 2 \sin \theta) \]
\[ Q = (-2 \cos \theta, -2 \sin \theta) \]

### Step 2: Find the Perpendicular Distances \( \alpha \) and \( \beta \)
The line equation is:
\[ x + y = 2 \]
The general formula for the perpendicular distance from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(x + y - 2 = 0\), we have \(A = 1\), \(B = 1\), and \(C = -2\).

#### Calculate \( \alpha \) (distance from point \(P\)):
Substituting \(P = (2 \cos \theta, 2 \sin \theta)\):
\[ \alpha = \frac{|1 \cdot (2 \cos \theta) + 1 \cdot (2 \sin \theta) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|2 \cos \theta + 2 \sin \theta - 2|}{\sqrt{2}} = \frac{2 |\cos \theta + \sin \theta - 1|}{\sqrt{2}} \]

#### Calculate \( \beta \) (distance from point \(Q\)):
Substituting \(Q = (-2 \cos \theta, -2 \sin \theta)\):
\[ \beta = \frac{|1 \cdot (-2 \cos \theta) + 1 \cdot (-2 \sin \theta) - 2|}{\sqrt{2}} = \frac{|-2 \cos \theta - 2 \sin \theta - 2|}{\sqrt{2}} = \frac{2 |-\cos \theta - \sin \theta - 1|}{\sqrt{2}} \]

### Step 3: Express \( \alpha \beta \)
Now we can express the product \( \alpha \beta \):
\[ \alpha \beta = \left(\frac{2 |\cos \theta + \sin \theta - 1|}{\sqrt{2}}\right) \left(\frac{2 |-\cos \theta - \sin \theta - 1|}{\sqrt{2}}\right) \]
\[ = \frac{4 |\cos \theta + \sin \theta - 1| \cdot |-\cos \theta - \sin \theta - 1|}{2} = 2 |\cos \theta + \sin \theta - 1| \cdot |-\cos \theta - \sin \theta - 1| \]

### Step 4: Simplify the Expression
Let \(x = \cos \theta + \sin \theta\). Then we can rewrite:
\[ \alpha \beta = 2 |x - 1| \cdot |-x - 1| = 2 |x - 1| \cdot |-(x + 1)| = 2 |x - 1| \cdot |x + 1| \]
This can be simplified to:
\[ \alpha \beta = 2 |(x - 1)(x + 1)| = 2 |x^2 - 1| \]

### Step 5: Find the Maximum Value
The maximum value of \(x = \cos \theta + \sin \theta\) occurs when \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{5\pi}{4}\):
\[ x = \sqrt{2} \quad \text{(maximum)} \]
Thus,
\[ x^2 = 2 \quad \text{and} \quad x^2 - 1 = 1 \]
So,
\[ \alpha \beta = 2 |1| = 2 \]

### Final Answer
The maximum value of \(\alpha \beta\) is:
\[ \boxed{2} \]