Question

If \[\left( {{m}_{i}},\dfrac{1}{{{m}_{i}}} \right),i=1,2,3,4\] are concyclic points then the value of \[{{m}_{1}}{{m}_{2}}{{m}_{3}}{{m}_{4}}\] is?
(A) 1
(B) -1
(C) 0
(D) \[\infty \]

Answer 1

Hint: So here in this question we have given that \[\left( {{m}_{i}},\dfrac{1}{{{m}_{i}}} \right),i=1,2,3,4\] and these are the concyclic points in geometry, a set of points are said to be concyclic if they lie on a common circle. All concyclic points are at the same distance from the center of the circle. Three points in the plane that do not all fall on a straight line are concyclic, but four or more such points in the plane are not necessarily concyclic. To solve the above equation, we have to know the equation of the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] here in this question we have to put $y=\dfrac{1}{x}$ then we will solve the equation so that we will get the value of \[{{m}_{1}}{{m}_{2}}{{m}_{3}}{{m}_{4}}\]

Complete step-by-step solution:
Given \[\left( {{m}_{i}},\dfrac{1}{{{m}_{i}}} \right),i=1,2,3,4\] are concyclic points means they lie on circle.
General equation for circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\].
Given general point is \[\left( x,\dfrac{1}{x} \right)\] . This point will satisfy the equation of circle.
\[\Rightarrow {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]
In the above equation we will put $y=\dfrac{1}{x}$ then we will get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}+2gx+2f\left( \dfrac{1}{x} \right)+c=0 \\
 & \Rightarrow {{x}^{4}}+2g{{x}^{3}}+c{{x}^{2}}+2fx+1=0 \\
\end{align}\]
From the above equation we will be compare it with the general form so that we can get the value of the \[{{m}_{1}},{{m}_{2}},{{m}_{3}},{{m}_{4}}\]
\[{{m}_{1}},{{m}_{2}},{{m}_{3}},{{m}_{4}}\] are the roots of the above equation.
As we know if an equation is \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0\] and roots of this equation are \[{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\]. Then \[{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}=\dfrac{e}{a}\].
\[\begin{align}
  & \Rightarrow {{m}_{1}}{{m}_{2}}{{m}_{3}}{{m}_{4}}=\dfrac{1}{1} \\
 & \Rightarrow {{m}_{1}}{{m}_{2}}{{m}_{3}}{{m}_{4}}=1 \\
\end{align}\]
Hence the correct option is (A) 1.

Note: Write the general point and general equation of circle. Then satisfy that point in the circle equation. Then solve that expression. In a general polynomial \[{{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+{{a}_{n-2}}{{x}^{n-2}}+....+{{a}_{1}}x+{{a}_{0}}=0\] and \[{{x}_{1}},{{x}_{2}},.....{{x}_{n-1}},{{x}_{n}}\] are roots.
Then multiplication of roots \[{{x}_{1}}{{x}_{2}}.....{{x}_{n-1}}{{x}_{n}}=\dfrac{{{\left( -1 \right)}^{n}}{{a}_{0}}}{{{a}_{n}}}\]
Be aware of the negative sign.