As easy as a, b, c, d?

All points are in the form $(x, 1/x)$ , i.e., for all points, $x^2 + \frac{1}{x^2} = 4^2$ , or $x^4 - 16x^2 + 1 = 0$ .

By Vieta, the product of all four roots $a, b, c, d$ is $\fbox{1}$

Let the centre of the circle be $(-g , -f).$ Now we have points $(a , \frac{1 }{ a }) , (b , \frac{1}{ b}) , (c ,\frac{ 1}{c} ) and (d ,\frac{1}{d} )$ which lies on a circle of radius of $4$ units . So this means that the points are at a distance of four units from the centre . Using distance formula , $\sqrt{[x-(-g)]^2 +[\frac {1}{x} - (-f)]^2 } = 4.$

Squaring both sides we get , $(x+g)^2 + (\frac{1}{x} + f)^2 =16.$

Now opening the brackets and multiplying throughout by $x^2$ we get ,

$x^4 + 2gx^3 + (g^2 + f^2 -16)x^2 + 2fx + 1 = 0$

Now since $a , b , c$ $and$ $d$ are the roots of this equation and by using Sir Vieta's formula we get ,

$\boxed {abcd = 1}$

Consider a circle (x-a)^2 + (y-b)^2 = r^2

put point (x,1/x) in circle . by Vieta's we get abcd = 1 independent of radius of circle

We have if $(a,\frac{1}{a})$ satisfies $x^2 + y^2 = 16$ then $(\frac{1}{a},a)$ also does too. WLOG let $b = \frac{1}{a}$ and $d = \frac{1}{c}$ . We get $abcd = a\frac{1}{a}c\frac{1}{c} = \boxed{1}$ .

If you set b=1/a c=-a d=-1/a you get 4 distinct points that have the same distance from the origin. You can find a in order to obtain a circle with radius 4 but you don't need it because abcd that is a(1/a)(-a)(-1/a) will be always equal to 1.