JEE Main 2022 — Differential Equations Question with Solution

Given a circle passing through $\left(0,-2\right)$ and $\left(0,2\right)$, so its equation will be,

$\left(\mathrm{x}-{\mathrm{x}}_1\right)\left(\mathrm{x}-{\mathrm{x}}_2\right)+\left(\mathrm{y}-y_1\right)\left(y-y_2\right)+{\lambda }_1\left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0,\left({\lambda }_1\in \mathrm{R}\right)$

On putting the given values we get,

$\Rightarrow \left(\mathrm{x}-0\right)\left(\mathrm{x}-0\right)+\left(\mathrm{y}-2\right)\left(y+2\right)+{\lambda }_1\left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& 1\\ 0& 2& 1\\ 0& -2& 1\end{array}\right|=0,\left({\lambda }_1\in \mathrm{R}\right)$

$\Rightarrow x^2+\left({\mathrm{y}}^2-4\right)+{\lambda }_1\left(4x\right)=0$ 

$\Rightarrow {\mathrm{x}}^2+\left({\mathrm{y}}^2-4\right)+\lambda \mathrm{x}=0,\left(\lambda \in \mathrm{R}\right)$

Now dividing both side by $x$ we get,

$\Rightarrow \frac{x^2+\left(y^2-4\right)}{x}+\lambda =0$

Now differentiating with respect to $x$

$\Rightarrow \frac{x\left[2x+2y·\frac{dy}{dx}\right]-\left[x^2+y^2-4\right]·1}{x^2}=0$

$\Rightarrow 2xy·\frac{dy}{dx}+\left(x^2-y^2+4\right)=0$