If `(m_(i),1/(m_(i))),m_(i)gt0,i=1,2,3,4` are four distinct points on a circle, show that `m_(1)m_(2)m_(3)m_(4)=1`.

Answer

To solve the problem, we need to show that if the points \((m_i, \frac{1}{m_i})\) for \(i = 1, 2, 3, 4\) are distinct points on a circle, then the product \(m_1 m_2 m_3 m_4 = 1\).

### Step-by-Step Solution:

1. **Equation of the Circle**:
We start with the general equation of a circle in the Cartesian plane:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g\), \(f\), and \(c\) are constants.

2. **Substituting Points**:
We substitute the points \((m_i, \frac{1}{m_i})\) into the circle's equation. For a specific point \((m, \frac{1}{m})\), we have:
\[ m^2 + \left(\frac{1}{m}\right)^2 + 2gm + 2f\left(\frac{1}{m}\right) + c = 0 \]

3. **Simplifying the Equation**:
We simplify the equation:
\[ m^2 + \frac{1}{m^2} + 2gm + \frac{2f}{m} + c = 0 \]
To eliminate the fraction, we multiply through by \(m^2\):
\[ m^4 + 1 + 2gm^3 + 2fm + cm^2 = 0 \]

4. **Rearranging the Equation**:
Rearranging gives us a polynomial in \(m\):
\[ m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0 \]
This is a quartic equation (degree 4).

5. **Roots of the Polynomial**:
Let \(m_1, m_2, m_3, m_4\) be the roots of the polynomial. According to Vieta's formulas, the product of the roots of a polynomial \(ax^n + bx^{n-1} + ... + z = 0\) is given by:
\[ \text{Product of roots} = (-1)^n \frac{z}{a} \]
Here, \(n = 4\) (degree of the polynomial), \(z = 1\) (constant term), and \(a = 1\) (coefficient of \(m^4\)).

6. **Calculating the Product**:
Therefore, we have:
\[ m_1 m_2 m_3 m_4 = (-1)^4 \frac{1}{1} = 1 \]

7. **Conclusion**:
Thus, we conclude that:
\[ m_1 m_2 m_3 m_4 = 1 \]