Question

If $\begin{pmatrix} a, & \frac{1}{a} \end{pmatrix}$, $\begin{pmatrix} b, & \frac{1}{b} \end{pmatrix}$, $\begin{pmatrix} c, & \frac{1}{c} \end{pmatrix}$ and $\begin{pmatrix} d, & \frac{1}{d} \end{pmatrix}$ are four distinct points on a circle of radius 4 units then, abcd is equal to

• A. 4
• B. 1/4
• C. 1
• D. 16

Answer 1

Answer: (C)

1

Answer 2

The four distinct points are given in the form $(x, 1/x)$. Let these points be $P_1(a, 1/a)$, $P_2(b, 1/b)$, $P_3(c, 1/c)$, and $P_4(d, 1/d)$. These points lie on a circle of radius $R=4$ units. The general equation of a circle with center $(h, k)$ and radius $R$ is: $(x-h)^2 + (y-k)^2 = R^2$ Given $R=4$, the equation is: $(x-h)^2 + (y-k)^2 = 16$ Since the points $(a, 1/a)$, $(b, 1/b)$, $(c, 1/c)$, and $(d, 1/d)$ lie on this circle, their coordinates must satisfy the circle's equation. For any point $(x, 1/x)$ on the circle, we have: $(x-h)^2 + \left(\frac{1}{x}-k\right)^2 = 16$ Expanding this equation: $x^2 - 2hx + h^2 + \frac{1}{x^2} - \frac{2k}{x} + k^2 = 16$ Multiplying by $x^2$ to eliminate fractions (note $x \neq 0$): $x^4 - 2hx^3 + h^2x^2 + 1 - 2kx + k^2x^2 = 16x^2$ Rearranging into a quartic equation: $x^4 - 2hx^3 + (h^2 + k^2 - 16)x^2 - 2kx + 1 = 0$ The roots of this quartic equation are $a, b, c,$ and $d$. By Vieta's formulas, the product of the roots of $Ax^4 + Bx^3 + Cx^2 + Dx + E = 0$ is $E/A$. In this equation, $A=1$ and $E=1$. Therefore, the product of the roots $abcd = 1/1 = 1$.