Circles are drawn on chords of the rectangular hyperbola $x y = 4$ parallel to the line $y = x$ as diameters.All such circles pass through two fixed points whose coordinates are

(2, 2)

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(2, - 2)

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(- 2, 2)

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(- 2, - 2)

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Solution

The correct options are
B
$(- 2, - 2)$

C $(2, 2)$
Given:Rectangular hyperbola $x y = 4 = c^{2}$
$\Rightarrow c^{2} = 4$
$\Rightarrow c = 2$
Let $P$ and $Q$ be the end points on the Rectangular hyperbola where $P (2 t_{1}, \frac{2}{t_{1}})$ and $Q (2 t_{2}, \frac{2}{t_{2}})$
Using the end points of diameter the equation of the circle

$C : (x - 2 t_{1}) (x - 2 t_{2}) + (y - \frac{2}{t_{1}}) (y - \frac{2}{t_{2}}) = 1$ .......... $(1)$

Now,Slope of the $P Q = \frac{\frac{2}{t_{2}} - \frac{2}{t_{1}}}{2 t_{2} - 2 t_{1}}$

$= \frac{2 (\frac{1}{t_{2}} - \frac{1}{t_{1}})}{2 (t_{2} - t_{1})}$

$= \frac{\frac{t_{1} - t_{2}}{t_{1} t_{2}}}{(t_{2} - t_{1})}$

$= \frac{\frac{t_{1} - t_{2}}{t_{1} t_{2}}}{(t_{2} - t_{1})}$

$= \frac{- 1}{t_{1} t_{2}}$

Hence $P Q$ is the diameter for circle and it is parallel to the line $y = x$

Slope of $P Q =$ Slope of the line $y = x$

$\Rightarrow \frac{- 1}{t_{1} t_{2}} = 1$

$\Rightarrow t_{1} t_{2} = - 1$

$(1) \Rightarrow (x - 2 t_{1}) (x - 2 t_{2}) + (y - \frac{2}{t_{1}}) (y - \frac{2}{t_{2}}) = 1$

$\Rightarrow x (x - 2 t_{2}) - 2 t_{1} (x - 2 t_{2}) + y (y - \frac{2}{t_{2}}) - \frac{2}{t_{1}} (y - \frac{2}{t_{2}}) = 1$
$\Rightarrow x^{2} - 2 x t_{2} - 2 x t_{1} + 4 t_{1} t_{2} + y^{2} - \frac{2 y}{t_{2}} - \frac{2 y}{t_{1}} + \frac{4}{t_{1} t_{2}} = 1$

$\Rightarrow x^{2} - 2 x t_{2} - 2 x t_{1} + 4 \times - 1 + y^{2} - \frac{2 y}{t_{2}} - \frac{2 y}{t_{1}} + \frac{4}{\times - 1} = 1$ using $t_{1} t_{2} = - 1$

$\Rightarrow x^{2} + y^{2} - 2 x (t_{2} + t_{1}) - 4 - 2 y (\frac{1}{t_{2}} + \frac{1}{t_{1}}) - 4 = 1$

$\Rightarrow x^{2} + y^{2} - 8 - 2 x (t_{2} + t_{1}) - 2 y (\frac{t_{1} + t_{2}}{t_{1} t_{2}}) = 1$

$\Rightarrow x^{2} + y^{2} - 8 - 2 x (t_{2} + t_{1}) - 2 y (\frac{t_{1} + t_{2}}{- 1}) = 1$ using $t_{1} t_{2} = - 1$

$\Rightarrow x^{2} + y^{2} - 8 - 2 x (t_{2} + t_{1}) + 2 y (t_{1} + t_{2}) = 1$

$\Rightarrow x^{2} + y^{2} - 8 + (2 y - 2 x) (t_{2} + t_{1}) = 1$ is of the form $C + λ L$

where $C = x^{2} + y^{2} - 8 = 0$ is the equation of a circle.
and $L = 2 y - 2 x = 0$ is the equation of a line.
$\Rightarrow y - x = 0$ or $x = y$

Substituting $x = y$ in the equation $x^{2} + y^{2} - 8 = 0$ we get
$\Rightarrow 2 x^{2} - 8 = 0$

$\Rightarrow 2 (x^{2} - 4) = 0$

$\Rightarrow (x - 2) (x + 2) = 0$

$∴ x = 2, - 2$

$\Rightarrow y = 2, - 2$ since $x = y$

Hence the coordinates of the fixed points are $(2, 2)$ and $(- 2, - 2)$

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