For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$, suppose the points $\left(x_1,…
For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$, suppose the points $\left(x_1, \frac{1}{x_1}\right),\left(x_2, \frac{1}{x_2}\right),\left(x_3, \frac{1}{x_3}\right)$ and $\left(x_4, \frac{1}{x_4}\right)$ lie on the boundary of a circle of radius 4 . Then, the value of $x_1 x_2 x_3 x_4$ is
$1$
$2$
$4$
$\frac{1}{4}$
Solution
Let the equation of the circle be $x^2+y^2+2 g x+2 f y+c=0$
$\left(x_i, \frac{1}{x_i}\right)$ lies on the above circle.
$\therefore \quad x_i^2+\frac{1}{x_i^2}+2 g x_i+\frac{2 f}{x_i}+C=0$
$\begin{aligned} & \Rightarrow \quad x_i^4+2 g x_i^3+c x_i^2+2 f x_i+1=0 \\ & \therefore \quad x_1 x_2 x_3 x_4=1\end{aligned}$