Mathematics · Differential Equations

JEE Main 2024 — 5 April, Shift 2 — Question 2

The differential equation of the family of circles passing the origin and having center at the line $\mathrm{y}=\mathrm{x}$ is :

Option A:
$\left(x^{2}-y^{2}+2 x y\right) d x=\left(x^{2}-y^{2}+2 x y\right) d y$
Option B:
$\left(x^{2}+y^{2}+2 x y\right) d x=\left(x^{2}+y^{2}-2 x y\right) d y$
Option C:
$\left(x^{2}-y^{2}+2 x y\right) d x=\left(x^{2}-y^{2}-2 x y\right) d y$
Correct
Option D:
$\left(x^{2}+y^{2}-2 x y\right) d x=\left(x^{2}+y^{2}+2 x y\right) d y$

Answer: C

$C \equiv x^{2}+y^{2}+g x+g y=0$

$2 x+2 y y^{\prime}+g+g y^{\prime}=0$

$g=-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)$

Put in (1)

$x^{2}+y^{2}-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)(x+y)=0$

$\left(x^{2}-y^{2}-2 x y\right) y^{\prime}=x^{2}-y^{2}+2 x y$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.