If `(m_(i),1//m_(i)), i=1,2,3,4` are concyclic points, then the value of `m_(1)m_(2)m_(3)m_(4)" is"`

Options

A. 1
B. `-1`
C. 0
D. `oo`

Answer

To solve the problem, we need to determine the value of \( m_1 m_2 m_3 m_4 \) given that the points \( (m_i, \frac{1}{m_i}) \) for \( i=1,2,3,4 \) are concyclic.

### Step-by-Step Solution:

1. **Understanding the Points**: The points given are \( (m_1, \frac{1}{m_1}), (m_2, \frac{1}{m_2}), (m_3, \frac{1}{m_3}), (m_4, \frac{1}{m_4}) \). These points lie on a circle, meaning they satisfy the equation of a circle.

2. **Equation of the Circle**: The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( g, f, c \) are constants.

3. **Substituting the Points into the Circle's Equation**: For each point \( (m_i, \frac{1}{m_i}) \), we substitute \( x = m_i \) and \( y = \frac{1}{m_i} \) into the circle's equation:
\[ m_i^2 + \left(\frac{1}{m_i}\right)^2 + 2g m_i + 2f \left(\frac{1}{m_i}\right) + c = 0 \]

4. **Multiplying Through by \( m_i^2 \)**: To eliminate the fraction, multiply the entire equation by \( m_i^2 \):
\[ m_i^4 + 1 + 2g m_i^3 + 2f m_i + c m_i^2 = 0 \]

5. **Rearranging the Equation**: Rearranging gives us:
\[ m_i^4 + 2g m_i^3 + c m_i^2 + 2f m_i + 1 = 0 \]

6. **Forming a Polynomial**: This is a polynomial of degree 4 in \( m \) with roots \( m_1, m_2, m_3, m_4 \).

7. **Using Vieta's Formulas**: According to Vieta's formulas, the product of the roots of a polynomial \( ax^4 + bx^3 + cx^2 + dx + e = 0 \) is given by:
\[ \text{Product of roots} = \frac{e}{a} \]
Here, \( a = 1 \) (coefficient of \( m^4 \)) and \( e = 1 \) (constant term).

8. **Calculating the Product**: Therefore, we have:
\[ m_1 m_2 m_3 m_4 = \frac{1}{1} = 1 \]

### Final Answer:
The value of \( m_1 m_2 m_3 m_4 \) is \( 1 \).