(a,(1/a)),(b,(1/b)),(c,(1/c)),(d,(1/d)) are four distinct points on a circle of radius is 4 units then abcd is equal to ?


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Question Number 31569 by momo last updated on 10/Mar/18

$(a, \frac{1}{a}), (b, \frac{1}{b}), (c, \frac{1}{c}), (d, \frac{1}{d}) a r e f o u r$ $d i s t i n c t p o i n t s o n a c i r c l e o f r a d i u s$ $i s 4 u n i t s t h e n a b c d i s e q u a l t o ?$
	Answered by MJS last updated on 10/Mar/18

	$center = (0; 0) :$ $x^{2} + y^{2} - 16 = 0$ $y = \frac{1}{x}$ $\Rightarrow x^{2} + \frac{1}{x^{2}} - 16 = 0$ $x^{4} - 16 x^{2} + 1 = 0$ $x = \sqrt{t}$ $t^{2} - 16 t + 1 = 0$ $t = 8 \pm \sqrt{64 - 1} = 8 \pm 3 \sqrt{7}$ $x_{1} = - \sqrt{8 + 3 \sqrt{7}}$ $x_{2} = - \sqrt{8 - 3 \sqrt{7}}$ $x_{3} = \sqrt{8 - 3 \sqrt{7}}$ $x_{4} = \sqrt{8 + 3 \sqrt{7}}$ $a b c d = x_{1} x_{2} x_{3} x_{4} = 1$ $center = (m; n)$ ${(x - m)}^{2} + {(y - n)}^{2} = r^{2}$ $y = \frac{1}{x}$ ${(x - m)}^{2} + {(\frac{1}{x} - n)}^{2} - r^{2} = 0$ $x^{2} - 2 m x + m^{2} + \frac{1}{x^{2}} - \frac{2 n}{x} + n^{2} - r^{2} = 0$ $x^{4} - 2 {m x}^{3} + (m^{2} + n^{2} - r^{2}) x^{2} - 2 n x + 1 = 0$ $\Rightarrow x_{1} x_{2} x_{3} x_{4} = 1$ $for x_{i} \in C$
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