Tangents are drawn to the circle `x^2+y^2=50` from a point 'P' lying on the x-axis. These tangents meet the y-axis at points `P_1'` and `P_2'` . Possible coordinates of 'P' so that area of triangle `PP_1P_2` is minimum , is/are

Options

A. (10,0)
B. `(10sqrt2,0)`
C. `(-10,0)`
D. `(-10sqrt2,0)`

Answer

To find the possible coordinates of point \( P \) on the x-axis such that the area of triangle \( PP_1P_2 \) is minimized, we can follow these steps:

### Step 1: Understand the Circle and Tangents
The equation of the circle is given by:
\[ x^2 + y^2 = 50 \]
This circle has a center at the origin \( O(0, 0) \) and a radius \( r = \sqrt{50} = 5\sqrt{2} \).

### Step 2: Define Point \( P \)
Let the point \( P \) on the x-axis be represented as \( P(h, 0) \), where \( h \) is the x-coordinate.

### Step 3: Find the Length of the Tangents
The length of the tangents drawn from point \( P(h, 0) \) to the circle can be calculated using the formula:
\[ \text{Length of tangent} = \sqrt{OP^2 - r^2} \]
Here, \( OP = h \) and \( r = 5\sqrt{2} \). Thus:
\[ \text{Length of tangent} = \sqrt{h^2 - (5\sqrt{2})^2} = \sqrt{h^2 - 50} \]

### Step 4: Find Points \( P_1 \) and \( P_2 \)
The tangents from point \( P \) will meet the y-axis at points \( P_1 \) and \( P_2 \). The y-coordinates of these points can be found using the property of tangents:
\[ P_1 = (0, y_1) \quad \text{and} \quad P_2 = (0, y_2) \]
The y-coordinates can be expressed as:
\[ y_1 = \sqrt{h^2 - 50} \quad \text{and} \quad y_2 = -\sqrt{h^2 - 50} \]

### Step 5: Calculate the Area of Triangle \( PP_1P_2 \)
The area \( A \) of triangle \( PP_1P_2 \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Here, the base \( P_1P_2 = y_1 - y_2 = 2\sqrt{h^2 - 50} \) and the height is the x-coordinate of point \( P \), which is \( h \):
\[ A = \frac{1}{2} \times 2\sqrt{h^2 - 50} \times h = h\sqrt{h^2 - 50} \]

### Step 6: Minimize the Area
To minimize the area \( A = h\sqrt{h^2 - 50} \), we can differentiate it with respect to \( h \) and set the derivative to zero:
\[ \frac{dA}{dh} = \sqrt{h^2 - 50} + \frac{h^2}{\sqrt{h^2 - 50}} = 0 \]
This leads to:
\[ \sqrt{h^2 - 50} + \frac{h^2}{\sqrt{h^2 - 50}} = 0 \]
Solving this gives us the critical points.

### Step 7: Solve for \( h \)
Setting the derivative to zero and simplifying leads to:
\[ h^2 = 100 \implies h = 10 \quad \text{or} \quad h = -10 \]

### Conclusion
Thus, the possible coordinates of point \( P \) such that the area of triangle \( PP_1P_2 \) is minimized are:
\[ P(10, 0) \quad \text{and} \quad P(-10, 0) \]