Consider the circle `x^(2) + y^(2) = 9` . Let P by any point lying on the positive x-axis. Tangents are drawn from this point to the given circle, meeting the Y-axis at `P_(1) and P_(2)` respectively. Then find the coordinates of point 'P' so that the area of the `DeltaPP_(1)P_(2)` is minimum.

Answer

To solve the problem, we need to find the coordinates of point \( P \) on the positive x-axis such that the area of triangle \( \Delta P P_1 P_2 \) is minimized, where \( P_1 \) and \( P_2 \) are the points where the tangents from \( P \) touch the circle defined by the equation \( x^2 + y^2 = 9 \).

### Step-by-Step Solution:

1. **Identify the Circle and Point \( P \)**:
The equation of the circle is \( x^2 + y^2 = 9 \). This circle has a center at \( (0, 0) \) and a radius \( r = 3 \). Let point \( P \) be \( (x, 0) \) on the positive x-axis.

2. **Use the Tangent Length Formula**:
The length of the tangent from a point \( (x_1, y_1) \) to a circle with center \( (0, 0) \) and radius \( r \) is given by:
\[ L = \sqrt{x_1^2 + y_1^2 - r^2} \]
For point \( P = (x, 0) \):
\[ L = \sqrt{x^2 + 0^2 - 9} = \sqrt{x^2 - 9} \]

3. **Coordinates of Points \( P_1 \) and \( P_2 \)**:
The points \( P_1 \) and \( P_2 \) where the tangents touch the circle can be found using the angle \( \alpha \) that the radius makes with the x-axis. The coordinates of \( P_1 \) and \( P_2 \) can be expressed as:
\[ P_1 = (0, y_1) \quad \text{and} \quad P_2 = (0, y_2) \]
where \( y_1 = 3 \sin(\alpha) \) and \( y_2 = -3 \sin(\alpha) \).

4. **Area of Triangle \( \Delta P P_1 P_2 \)**:
The area \( A \) of triangle \( \Delta P P_1 P_2 \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Here, the base \( P_1P_2 = 2 \times y_1 = 6 \sin(\alpha) \) and the height from \( P \) to the line \( P_1P_2 \) is \( OP = 3 \sec(\alpha) \). Therefore:
\[ A = \frac{1}{2} \times 6 \sin(\alpha) \times 3 \sec(\alpha) = 9 \tan(\alpha) \]

5. **Minimizing the Area**:
To minimize the area \( A = 9 \tan(\alpha) \), we need to find the maximum value of \( \tan(\alpha) \). The maximum value of \( \tan(\alpha) \) occurs when \( \alpha = \frac{\pi}{4} \), where \( \tan(\frac{\pi}{4}) = 1 \).

6. **Finding Coordinates of Point \( P \)**:
Substituting \( \alpha = \frac{\pi}{4} \):
\[ OP = 3 \sec\left(\frac{\pi}{4}\right) = 3 \cdot \sqrt{2} \]
Therefore, the coordinates of point \( P \) are:
\[ P = (3\sqrt{2}, 0) \]

### Final Answer:
The coordinates of point \( P \) are \( (3\sqrt{2}, 0) \).