The differential equation representing all possible curves that cut each member of the family of circles
$x^{2} + y^{2} - 2 C x = 0$ (C is a parameter) at right angle, is

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$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y}$

$\frac{d y}{d x} = \frac{2 x y}{x^{2} + y^{2}}$

$\frac{d y}{d x} = \frac{2 x y}{x^{2} - y^{2}}$

$\frac{d y}{d x} = \frac{x^{2} - y^{2}}{2 x y}$

answer is B.

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Detailed Solution

Here, we have to find the orthogonal trajectories of the family of circles

$x^{2} + y^{2} - 2 C x = 0$ (i)

Differentiating (i) w.r.t. x, we get

$2 x + 2 y \frac{d y}{d x} - 2 C = 0 \Rightarrow C = x + y \frac{d y}{d x}$ …(ii)

From (i) and (ii), we obtain

$x^{2} + y^{2} - 2 x (x + y \frac{d y}{d x}) = 0$ [By eliminating C]

$\Rightarrow y^{2} - x^{2} - 2 x y \frac{d y}{d x} = 0 \Rightarrow y^{2} - x^{2} = 2 x y \frac{d y}{d x}$ ….(iii)

This is the differential equation representing the given family of circles. To find the differential equation of the orthogonal trajectories, we replace $\frac{d y}{d x} by - \frac{d x}{d y}$ in equation (iii) Thus, the differential equation representing the orthogonal trajectories is

$y^{2} - x^{2} = - 2 x y \frac{d x}{d y} or, \frac{d y}{d x} = \frac{2 x y}{x^{2} - y^{2}}$