Q40 / 90 · JEE Main 2024 (05 Apr Shift 2) · Mathematics

The differential equation of the family of circles passing through the origin and having centre at the line $y=x$ is :

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

Solution

$\begin{aligned} & C \equiv x^2+y^2+g x+g y=0....(1)\\ & 2 x+2 y y^{\prime}+g+g y^{\prime}=0 \\ & g=-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right) \end{aligned}$ Put in (1) $\begin{aligned} & x^2+y^2-\left(\frac{2 x+2 y y^{\prime}}{1+y^{\prime}}\right)(x+y)=0 \\ & \left(x^2-y^2-2 x y\right) y^{\prime}=x^2-y^2+2 x y \end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions