AB is a diameter of the circle `x^(2)+y^(2)=4`. If `p_(1)` and `p_(2)` be the lengths of perpendiculars from A and B on the line `x+y=1`, then maximum value of `p_(1)p_(2)` is:

Options

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

Answer

To solve the problem step by step, we will follow the given information and derive the required maximum value of \( p_1 p_2 \).

### Step 1: Identify the Circle and Points A and B
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This circle has a center at \( (0, 0) \) and a radius \( r = 2 \). The diameter \( AB \) can be represented by points on the circle. We can express points \( A \) and \( B \) in terms of a parameter \( t \):
\[ A = (2 \cos t, 2 \sin t) \]
\[ B = (2 \cos(t + \pi), 2 \sin(t + \pi)) = (-2 \cos t, -2 \sin t) \]

### Step 2: Find the Perpendicular Distances \( p_1 \) and \( p_2 \)
The line equation is given by:
\[ x + y = 1 \]
We can rewrite it in the standard form:
\[ x + y - 1 = 0 \]
The formula for the distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

#### For Point A:
Using \( A = (2 \cos t, 2 \sin t) \):
\[ p_1 = \frac{|(1)(2 \cos t) + (1)(2 \sin t) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2 \cos t + 2 \sin t - 1|}{\sqrt{2}} = \frac{|2(\cos t + \sin t) - 1|}{\sqrt{2}} \]

#### For Point B:
Using \( B = (-2 \cos t, -2 \sin t) \):
\[ p_2 = \frac{|(1)(-2 \cos t) + (1)(-2 \sin t) - 1|}{\sqrt{2}} = \frac{|-2 \cos t - 2 \sin t - 1|}{\sqrt{2}} = \frac{|-(2(\cos t + \sin t) + 1)|}{\sqrt{2}} = \frac{2(\cos t + \sin t) + 1}{\sqrt{2}} \]

### Step 3: Calculate \( p_1 p_2 \)
Now, we can find the product \( p_1 p_2 \):
\[ p_1 p_2 = \left(\frac{|2(\cos t + \sin t) - 1|}{\sqrt{2}}\right) \left(\frac{2(\cos t + \sin t) + 1}{\sqrt{2}}\right) \]
\[ = \frac{|(2(\cos t + \sin t) - 1)(2(\cos t + \sin t) + 1)|}{2} \]
Let \( z = \cos t + \sin t \):
\[ p_1 p_2 = \frac{|(2z - 1)(2z + 1)|}{2} = \frac{|4z^2 - 1|}{2} \]

### Step 4: Maximize \( p_1 p_2 \)
To maximize \( |4z^2 - 1| \), we need to find the maximum value of \( z = \cos t + \sin t \). The maximum value of \( z \) occurs when:
\[ z = \sqrt{2} \quad \text{(when \( t = \frac{\pi}{4} \))} \]
Thus, we have:
\[ 4z^2 = 4(\sqrt{2})^2 = 8 \]
So:
\[ |4z^2 - 1| = |8 - 1| = 7 \]
Finally, substituting back:
\[ p_1 p_2 = \frac{7}{2} \]

### Conclusion
The maximum value of \( p_1 p_2 \) is:
\[ \boxed{\frac{7}{2}} \]