Circles are drawn on chords of the rectangula | vedclass

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(D) Let the points on the hyperbola $xy = c^2$ be $P(ct_1, c/t_1)$ and $Q(ct_2, c/t_2)$.Since the chord $PQ$ is parallel to $y = x$,its slope is $1$.T

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both $(A)$ and $(B)$

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(D) Let the points on the hyperbola $xy = c^2$ be $P(ct_1, c/t_1)$ and $Q(ct_2, c/t_2)$.Since the chord $PQ$ is parallel to $y = x$,its slope is $1$.Thus,$\frac{c/t_2 - c/t_1}{ct_2 - ct_1} = 1$ $\Rightarrow \frac{c(t_1 - t_2)}{t_1t_2} \cdot \frac{1}{c(t_2 - t_1)} = 1$ $\Rightarrow -\frac{1}{t_1t_2} = 1$ $\Rightarrow t_1t_2 = -1$.The equation of the circle with diameter $PQ$ is $(x - ct_1)(x - ct_2) + (y - c/t_1)(y - c/t_2) = 0$.Expanding this,we get $x^2 - c(t_1 + t_2)x + c^2t_1t_2 + y^2 - c(1/t_1 + 1/t_2)y + c^2/(t_1t_2) = 0$.Substituting $t_1t_2 = -1$,we get $x^2 + y^2 - c(t_1 + t_2)x - c\left(\frac{t_1 + t_2}{t_1t_2}\right)y - c^2 - c^2 = 0$.Since $t_1t_2 = -1$,this simplifies to $x^2 + y^2 - c(t_1 + t_2)x + c(t_1 + t_2)y - 2c^2 = 0$.Rearranging,we get $(x^2 + y^2 - 2c^2) - c(t_1 + t_2)(x - y) = 0$.For this to pass through fixed points,the terms must satisfy $x^2 + y^2 - 2c^2 = 0$ and $x - y = 0$.Substituting $y = x$ into the first equation,$2x^2 = 2c^2$ $\Rightarrow x^2 = c^2$ $\Rightarrow x = \pm c$.Thus,the fixed points are $(c, c)$ and $(-c, -c)$.

Circles are drawn on chords of the rectangular hyperbola $xy = c^2$ parallel to the line $y = x$ as diameters. All such circles pass through two fixed points whose coordinates are:

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