If a circle and the rectangular hyperbola $x y = c^{2}$ meet in the four points $t_{1}, t_{2}, t_{3}$ and $t_{4}$ then

t_{1} t_{2} t_{3} t_{4} = 1

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The arthmetic mean of the four points bisects the distance between the centres of the two curves

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The geometrical mean of the four points bisects the distance between the centres of the two curves

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The centre of the circle through the points

t_{1}, t_{2}

and

t_{3}

is:

{\frac{c}{2} (t_{1} + t_{2} + t_{3} + \frac{1}{t_{1} t_{2} t_{3}}), \frac{c}{2} (\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + t_{1} t_{2} t_{3})}

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Solution

The correct options are
A $t_{1} t_{2} t_{3} t_{4} = 1$
B The arthmetic mean of the four points bisects the distance between the centres of the two curves
C The geometrical mean of the four points bisects the distance between the centres of the two curves
D The centre of the circle through the points $t_{1}, t_{2}$ and $t_{3}$ is: ${\frac{c}{2} (t_{1} + t_{2} + t_{3} + \frac{1}{t_{1} t_{2} t_{3}}), \frac{c}{2} (\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + t_{1} t_{2} t_{3})}$
A rectangular hyperbola and a circle meet in four points. The mean of these four points is the middle point of the centres of the hyperbola and that of the circle.
Let the rectangular hyperbola be $x y = c^{2}$ and the equation of the circle be $x^{2} + y^{2} + 2 g c t + 2 f y + k = 0$ . Any point on the hyperbola is $(c p, \frac{c}{t})$ . If it lies on the circle, then $c^{2} t^{2} + \frac{c^{2}}{t^{2}} + 2 g c t + 2 f \frac{c}{t} + k = 0$ .
$\Rightarrow c^{2} t^{4} + 2 g c t^{3} + k t^{2} + 2 f c t + c^{2} = 0$ .
This is fourth degree equation in $t$ , which has four roots. Hence the circle and the hyperbola intersect in four points. If $t_{1}, t_{2}, t_{3}, t_{4}$ are the roots of this equation, then
$t_{1} + t_{2} + t_{3} + t_{4} = \frac{- 2 g c}{c^{2}} = \frac{- 2 g}{c}$
$\Rightarrow c t_{1} + c t_{2} + c t_{3} + c t_{4} = 2 g$
$\Rightarrow x_{1} + x_{2} + x_{3} + \frac{x_{4}}{4} = \frac{- g}{2}$
Also $\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + \frac{1}{t_{4}}$
$\frac{\sum t_{1} t_{2} t_{3}}{t_{1} t_{2} t_{3} t_{4}} = \frac{- 2 f c}{c^{2}} = \frac{- 2 f}{c}$
$\frac{c}{t_{1}} + \frac{c}{t_{2}} + \frac{c}{t_{3}} + \frac{c}{t_{4}} = - 2 f$
$\Rightarrow y_{1} + y_{2} + y_{3} + \frac{y_{4}}{4} = \frac{- f}{2}$ .
Hence the mean of the four points is $(\frac{- g}{2}, \frac{- f}{2})$ which is the mid-point of the centre of the hyperbola and that of the circle.
Hence, if a circle and the rectangular hyperbola $x y = c^{2}$ meet in the four points $t_{1}, t_{2}, t_{3}, t_{4}$ , then
$(a) t_{1} t_{2} t_{3} t_{4} = 1$
$(b)$ The centre of the mean position of the four points bisects the distance between the centre of the two curves.
$(c)$ The centre of the circle through the points $t_{1}, t_{2}, t_{3}$ is ${\frac{c}{2} (t_{1} + t_{2} + t_{3} + \frac{1}{t_{1} t_{2} t_{3}}), \frac{c}{2} (\frac{1}{t_{1}} + \frac{1}{t_{2}} + \frac{1}{t_{3}} + t_{1} t_{2} t_{3})}$ .

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