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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

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

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

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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

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