JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Differential Equations

The differential equation of the family of circles passing through the origin and having centre at the line

y = x

Answer: (B)

Solution

A circle through the origin with centre on

y = x

has the form

x^{2} + y^{2} - 2 g x - 2 g y = 0

(centre

(g, g)

, radius through the origin automatically).
Differentiating:

2 x + 2 y y^{'} - 2 g - 2 g y^{'} = 0 ⟹ g = \frac{x + y y ^{'}}{1 + y ^{'}}

.
Substitute back into

x^{2} + y^{2} = 2 g (x + y)

(x^{2} + y^{2}) (1 + y^{'}) = 2 (x + y y^{'}) (x + y)

Expanding both sides and collecting terms in

y^{'}

(x^{2} - y^{2} + 2 x y) d x = (x^{2} - y^{2} - 2 x y) d y

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option