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

$$a{x^4} + b{x^3} + c{x^2} + dx + c = 0 \\\ \alpha \cdot \beta \cdot \gamma \cdot \delta = \dfrac{c}{a} \;$$