Question

\[f\left( {{m_i},\dfrac{1}{{{m_i}}}} \right),i = 1,2,3,4\] are four distinct points on the circle with centre origin, then value of \[{m_1}{m_2}{m_3}{m_4}\] is equal to
A. \[0\]
B. \[ - 1\]
C. \[1\]
D. \[ - a\]

Answer 1

Hint: Here, we have four distinct points on the circle with centre origin. A circle is the set of all points in a plane that are equidistant from a given point called the centre of the circle. The point lie on a circle equation is
\[
  a{x^4} + b{x^3} + c{x^2} + dx + c = 0 \\
  \alpha \cdot \beta \cdot \gamma \cdot \delta = \dfrac{c}{a} \;
\]

Complete step-by-step answer:
In the given problem,
Let \[f\left( {{m_i},\dfrac{1}{{{m_i}}}} \right)\] be the function of point \[f\left( {x,y} \right)\] , where, \[i = 1,2,3,4\] .
Let the four distinct points are \[{m_1},{m_2},{m_3},{m_4}\]
Let us consider \[f\left( {{m_i},\dfrac{1}{{{m_i}}}} \right)\] as \[f\left( {x,y} \right)\]
Let, \[x = {m_i}\] .
so, \[y = \dfrac{1}{{{m_i}}} = \dfrac{1}{x}\] .
The points lie on a circle, \[{x^2} + {y^2} + 2gx + 2fy + c = 0\] .
By substituting, \[y = \dfrac{1}{x}\] into the equation, we get
 \[{x^2} + {\left( {\dfrac{1}{x}} \right)^2} + 2gx + 2f\left( {\dfrac{1}{x}} \right) + c = 0\]
By simplifying, we get
 \[{x^2} + \dfrac{1}{{{x^2}}} + 2gx + \dfrac{{2f}}{x} + c = 0\]
Take LCM on the above equation, we get
 \[\dfrac{{{x^4} + 1 + 2g{x^3} + 2fx + c{x^2}}}{{{x^2}}} = 0\]
By perform multiplication on both sides by \[{x^2}\] , we get
 \[{x^4} + 2g{x^3} + c{x^2} + 2fx + 1 = 0 \to (1)\]
Let \[f\left( {{m_i},\dfrac{1}{{{m_i}}}} \right)\] for root of equation for \[i = 1,2,3,4\]
By substitute \[f\left( {{m_i},\dfrac{1}{{{m_i}}}} \right)\] in equation \[(1)\] .
 \[\left( 1 \right) \Rightarrow {m_i}^4 + 2g{m_i}^3 + c{m_i}^2 + 2f{m_i} + 1 = 0\] , Where, \[{m_i}\] are roots of the equation for \[i = 1,2,3,4\]
On comparing the above equation with root of the equation
\[
  a{x^4} + b{x^3} + c{x^2} + dx + c = 0 \\
  \alpha \cdot \beta \cdot \gamma \cdot \delta = \dfrac{c}{a} \;
\]
Substitute the given points and equations into the root of the equation, we get
Therefore, The value of
 \[
  {m_i}^4 + 2g{m_i}^3 + c{m_i}^2 + 2f{m_i} + 1 = 0 \\
  {m_1} \cdot {m_2} \cdot {m_3} \cdot {m_4} = \dfrac{1}{1} = 1 \;
\]
 \[{m_1} \cdot {m_2} \cdot {m_3} \cdot {m_4} = 1\]
Therefore, The value of \[{m_1}{m_2}{m_3}{m_4}\] is equal to \[1\] .
Final answer is option(c) \[1\] .
So, the correct answer is “Option C”.

Note: Here, we need to solve this problem by the root of the equation and it has four distinct points \[{m_1}\;{m_2}\;{m_3}\;{m_4}\] of the centre of origin by substitute the values to the equation
\[
  a{x^4} + b{x^3} + c{x^2} + dx + c = 0 \\
  \alpha \cdot \beta \cdot \gamma \cdot \delta = \dfrac{c}{a} \;
\]