SATHEE: Circle 1 Question 20

Circle 1 Question 20

20. If $(m_{i}, 1 / m_{i}), m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .

$(1989, 2 M)$

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Solution:

Let the points $m_{i}, \frac{1}{m_{i}}; i = 1, 2, 3, 4$ lie on a circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Then, $m_{i}^{2} + \frac{1}{m_{i}^{2}} + 2 g m_{i} + \frac{2 f}{m_{i}} + c = 0$ ;

Since, $m_{i}^{4} + 2 g m_{i}^{3} + c m_{i}^{2} + 2 f m_{i} + 1 = 0; i = 1, 2, 3, 4$ $\Rightarrow m_{1}, m_{2}, m_{3}$ and $m_{4}$ are the roots of the equation

$\begin{aligned} m^{4} + 2 g m^{3} + c m^{2} + 2 f m + 1 & = 0 \\ \Rightarrow & m_{1} m_{2} m_{3} m_{4} & = \frac{1}{1} = 1 \end{aligned}$

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Circle 1 Question 20

20. If (mi,1/mi),mi>0,i=1,2,3,4 are four distinct points on a circle, then show that m1m2m3m4=1.

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20. If $(m_{i}, 1 / m_{i}), m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .