Question

For different real non-zero numbers \(x_1,x_2,x_3\) and \(x_4\), suppose the points \[ \left(x_1,\frac{1}{x_1}\right),\left(x_2,\frac{1}{x_2}\right),\left(x_3,\frac{1}{x_3}\right) \textbf{ and } \left(x_4,\frac{1}{x_4}\right) \] lie on the boundary of a circle of radius \(4\). Then the value of \(x_1x_2x_3x_4\) is:

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(\dfrac{1}{4}\)

Hint:

When points are of the form \(\left(x,\frac{1}{x}\right)\), substitute \(y=\frac{1}{x}\) into the given curve equation and then multiply by \(x^2\) to form a polynomial in \(x\).

Answer 1

The Correct Option is A

Step 1: Write the general equation of the circle.
Let the centre of the circle be \[ (a,b). \] Since the radius is \(4\), the equation of the circle is \[ (x-a)^2+(y-b)^2=16. \] The given points are of the form \[ \left(x,\frac{1}{x}\right). \] So, \[ y=\frac{1}{x}. \]

Step 2: Substitute \(y=\frac{1}{x}\) in the circle equation.
Substituting, \[ (x-a)^2+\left(\frac{1}{x}-b\right)^2=16. \] Expanding, \[ x^2-2ax+a^2+\frac{1}{x^2}-\frac{2b}{x}+b^2=16. \]

Step 3: Convert into a polynomial equation.
Multiplying throughout by \(x^2\), we get \[ x^4-2ax^3+(a^2+b^2-16)x^2-2bx+1=0. \] The four different real non-zero numbers \[ x_1,x_2,x_3,x_4 \] are the four roots of this equation.

Step 4: Use product of roots.
For a fourth-degree equation \[ Ax^4+Bx^3+Cx^2+Dx+E=0, \] the product of roots is \[ \frac{E}{A}. \] Here, \[ A=1 \] and \[ E=1. \] Therefore, \[ x_1x_2x_3x_4=\frac{1}{1}=1. \]

Step 5: Final conclusion.
Hence, \[ \boxed{1}. \]