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Q. The differential equation of the family of circles passing through the points $(0,2)$ and $(0,-2)$ is

JEE MainJEE Main 2022Differential Equations

Solution:

Sol. Equation of circle passing through $(0,-2)$ and $(0,2)$ is
$x^2+\left(y^2-4\right)+\lambda x=0,(\lambda \in R)$
Divided by $x$ we get
$\frac{x^2+\left(y^2-4\right)}{x}+\lambda=0$
Differentiating with respect to $x$
$ \frac{x\left[2 x+2 y \cdot \frac{d y}{d x}\right]-\left[x^2+y^2-4\right] \cdot 1}{x^2}=0 $
$ \Rightarrow 2 x y \cdot \frac{d y}{d x}+\left(x^2-y^2+4\right)=0$