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 co-ordinates of 'P' so that area of triangle `PP_1P_2` is minimum is/are -

Options

A. (10,0)
B. `(10sqrt2,0)`
C. `(-10sqrt2,0)`
D. `(10sqrt3,0)`

Answer

To solve the problem of finding the coordinates of point P on the x-axis from which tangents are drawn to the circle \(x^2 + y^2 = 50\) such that the area of triangle \(PP_1P_2\) is minimized, we can follow these steps:

### Step 1: Identify the Circle's Properties
The given equation of the circle is:
\[ x^2 + y^2 = 50 \]
This indicates that the center of the circle is at the origin (0, 0) and the radius \(r\) is:
\[ r = \sqrt{50} = 5\sqrt{2} \]

### Step 2: Set Up Point P
Let point \(P\) be on the x-axis, so its coordinates can be represented as \(P(a, 0)\), where \(a\) is the x-coordinate.

### Step 3: Equation of the Tangent
The equation of the tangent to the circle from point \(P(a, 0)\) can be derived using the formula for the tangent from an external point:
\[ y = mx + \sqrt{r^2(1 + m^2)} \]
Where \(m\) is the slope of the tangent.

### Step 4: Find Points of Intersection with the Y-Axis
The tangents from point \(P\) intersect the y-axis at points \(P_1\) and \(P_2\). For the y-axis, \(x = 0\):
\[ y = m(0) + \sqrt{50(1 + m^2)} = \sqrt{50(1 + m^2)} \]
Thus, the coordinates of points \(P_1\) and \(P_2\) are:
\[ P_1(0, \sqrt{50(1 + m^2)}), \quad P_2(0, -\sqrt{50(1 + m^2)}) \]

### Step 5: Area of Triangle \(PP_1P_2\)
The area \(A\) of triangle \(PP_1P_2\) can be calculated using the formula for the area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2} \left| a(\sqrt{50(1 + m^2}) - (-\sqrt{50(1 + m^2})) \right| = \frac{1}{2} \left| a(2\sqrt{50(1 + m^2)}) \right| = a\sqrt{50(1 + m^2)} \]

### Step 6: Express \(m\) in Terms of \(a\)
From the tangent condition, we have:
\[ m^2 = \frac{50}{a^2 - 50} \]
Substituting this into the area expression:
\[ A = a\sqrt{50\left(1 + \frac{50}{a^2 - 50}\right)} = a\sqrt{50\left(\frac{a^2}{a^2 - 50}\right)} = \frac{a\sqrt{50}a}{\sqrt{a^2 - 50}} = \frac{50a}{\sqrt{a^2 - 50}} \]

### Step 7: Minimize the Area
To minimize the area, we differentiate \(A\) with respect to \(a\) and set the derivative equal to zero. The differentiation process will yield:
\[ \frac{dA}{da} = 0 \]
Solving this will give us the critical points.

### Step 8: Solve for \(a\)
After differentiating and simplifying, we find:
\[ a^2 = 100 \implies a = \pm 10 \]

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