Let C be the circle of radius unity centred at the origin. If two positive numbers `x_1 and x_2` are such that the line passing through `(x_1,-1) and (x_2, 1)` is tangent to C then `x_1*x_2`

Options

A. `x_(1)x_(2) = 1`
B. `x_(1)x_(2) =- 1`
C. `x_(1) +x_(2) = 1`
D. `4x_(1)x_(2)=1`

Explanation

To solve the problem, we need to find the product \( x_1 \cdot x_2 \) given that the line passing through the points \( (x_1, -1) \) and \( (x_2, 1) \) is tangent to the circle centered at the origin with a radius of 1.

### Step-by-Step Solution:

1. **Equation of the Circle**:
The equation of the circle \( C \) with radius 1 centered at the origin is given by:
\[ x^2 + y^2 = 1 \]

2. **Equation of the Line**:
The line passing through the points \( (x_1, -1) \) and \( (x_2, 1) \) can be expressed in the slope-intercept form. The slope \( m \) of the line connecting these two points is:
\[ m = \frac{1 - (-1)}{x_2 - x_1} = \frac{2}{x_2 - x_1} \]
Using point-slope form, the equation of the line can be written as:
\[ y - (-1) = m(x - x_1) \quad \text{or} \quad y + 1 = \frac{2}{x_2 - x_1}(x - x_1) \]
Rearranging gives:
\[ y = \frac{2}{x_2 - x_1}x - \frac{2x_1}{x_2 - x_1} - 1 \]

3. **Finding the Tangent Condition**:
For the line to be tangent to the circle, the perpendicular distance from the center of the circle (the origin) to the line must equal the radius (which is 1). The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line, we can rearrange it to the form \( Ax + By + C = 0 \):
\[ \frac{2}{x_2 - x_1}x - y - \left(-\frac{2x_1}{x_2 - x_1} - 1\right) = 0 \]
Here, \( A = \frac{2}{x_2 - x_1}, B = -1, C = -\left(-\frac{2x_1}{x_2 - x_1} - 1\right) \).

4. **Calculating the Distance**:
Substituting \( (0, 0) \) for \( (x_0, y_0) \):
\[ d = \frac{\left|0 + 0 + C\right|}{\sqrt{\left(\frac{2}{x_2 - x_1}\right)^2 + (-1)^2}} = 1 \]
This simplifies to:
\[ \frac{\left|-\frac{2x_1}{x_2 - x_1} - 1\right|}{\sqrt{\frac{4}{(x_2 - x_1)^2} + 1}} = 1 \]

5. **Squaring Both Sides**:
Squaring both sides and simplifying will lead us to the relationship between \( x_1 \) and \( x_2 \). After simplification, we will find:
\[ x_1 x_2 = 1 \]

### Conclusion:
Thus, the product \( x_1 \cdot x_2 \) is equal to 1.