It's all on the circle

Let equation of circle is $x^2 + y^2 + 2gx + 2fy + c = 0$
Since, $\left(a , \dfrac{1}{a}\right)$ lies on circle, we get
$\large a^2 + \dfrac{1}{a^2} + 2ga + 2f\left(\dfrac{1}{a}\right) + c = 0$ $\large a^4 + 1 + 2ga^3 + 2fa + ca^2 = 0$ Using Vieta's formula , $\large \displaystyle \sum abcd = 1$

On substituting the values $\left(a,\dfrac{1}{a}\right)$ , $\left(b,\dfrac{1}{b}\right)$ , $\left(c,\dfrac{1}{c}\right)$ and $\left(d,\dfrac{1}{d}\right)$ in the standard form of circle $(x^2 + y^2 + 2gx + 2fy + c = 0$ , we obtain four cyclic equations.

Note that if we swap, say $\left(a,\dfrac{1}{a}\right)$ with $\left(\dfrac{1}{a},a\right)$ the equations remain unchanged. Clearly, then $\large \frac{1}{a}$ = a.

Also, since the equations are cyclic $\large a =b = c=d$ . $\large 1$ & $\large -1$ satisfiy the required conditions.

If you think about the plot of the function 1/x that intersects a circumference whose center is the origin of the axes, their common points are 4. For the symmetry of 1/x respect (0,0) these points have equal and opposite x coordinate from two in two. Being also points of the circumference the problem is sym respect (0,0) so the same process can be repeated thinking in terms of the y coordinate. In particoular consider two points that have x coordinates (a,1/a) and (-a,-1/a), for symmetry the other two points have to have the y coordinate as a and -a , finally their x coordinates are 1/a and -1/a.

In the end the result of (a)(-1/a)(-a)(1/a) is 1