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, we need to find the maximum value of the product of the perpendicular distances (denoted as α and β) from the points P and Q on the circle \(x^2 + y^2 = 4\) to the line \(x + y = 2\).

### Step 1: Identify the Circle and Its Diameter
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This represents a circle with center at \((0, 0)\) and radius \(r = 2\). The points \(P\) and \(Q\) are endpoints of the diameter of this circle.

### Step 2: Parametrize the Points on the Circle
We can express the points \(P\) and \(Q\) on the circle using the parameter \(\theta\):
\[ P = (2 \cos \theta, 2 \sin \theta) \]
\[ Q = (2 \cos(\theta + \pi), 2 \sin(\theta + \pi) = (-2 \cos \theta, -2 \sin \theta) \]

### Step 3: Calculate the Perpendicular Distances
The line \(x + y = 2\) can be rewritten in the standard form as:
\[ x + y - 2 = 0 \]
The 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 \(C = -2\).

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

#### For Point Q:
Substituting \(Q = (-2 \cos \theta, -2 \sin \theta)\):
\[ \beta = \frac{|1(-2 \cos \theta) + 1(-2 \sin \theta) - 2|}{\sqrt{2}} = \frac{|-2 \cos \theta - 2 \sin \theta - 2|}{\sqrt{2}} = \frac{2 |-\cos \theta - \sin \theta - 1|}{\sqrt{2}} \]

### Step 4: Find the Product αβ
Now we need to find the product:
\[ \alpha \beta = \left(\frac{2 |\cos \theta + \sin \theta - 1|}{\sqrt{2}}\right) \left(\frac{2 |-\cos \theta - \sin \theta - 1|}{\sqrt{2}}\right) \]
\[ = \frac{4 |\cos \theta + \sin \theta - 1| |-\cos \theta - \sin \theta - 1|}{2} = 2 |\cos \theta + \sin \theta - 1| |-\cos \theta - \sin \theta - 1| \]

### Step 5: Simplify the Expression
Let \(u = \cos \theta + \sin \theta\). Then we can rewrite the product as:
\[ \alpha \beta = 2 |u - 1| |-(u + 1)| = 2 |u - 1| |u + 1| \]
This can be simplified to:
\[ \alpha \beta = 2 |u^2 - 1| = 2 |(\cos \theta + \sin \theta)^2 - 1| \]
Using the identity \((\cos \theta + \sin \theta)^2 = 1 + 2 \sin \theta \cos \theta = 1 + \sin 2\theta\):
\[ \alpha \beta = 2 |\sin 2\theta| \]

### Step 6: Find the Maximum Value
The maximum value of \(|\sin 2\theta|\) is \(1\). Therefore, the maximum value of \(\alpha \beta\) is:
\[ \text{Maximum value of } \alpha \beta = 2 \times 1 = 2 \]

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