The differential equation of the family of circles passing through...

Given a circle passing through $\left(0, - 2\right)$ and \left(0, 2\right), so its equation will be,</p><p>\left(x - {x}_{1}\right)\left(x - {x}_{2}\right) + \left(y - {y}_{1}\right)\left(y - {y}_{2}\right) + {\lambda}_{1}\left|\begin{cases}x & y & 1 \\ {x}_{1} & {y}_{1} & 1 \\ {x}_{2} & {y}_{2} & 1\end{cases}\right| = 0, \left({\lambda}_{1} \in R\right)

On putting the given values we get,

⇒ \left(x - 0\right)\left(x - 0\right) + \left(y - 2\right)\left(y + 2\right) + {\lambda}_{1}\left|\begin{cases}x & y & 1 \\ 0 & 2 & 1 \\ 0 & - 2 & 1\end{cases}\right| = 0, \left({\lambda}_{1} \in R\right)

$⇒ {x}^{2} + \left({y}^{2} - 4\right) + {\lambda}_{1}\left(4x\right) = 0$

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

Now dividing both side by $x$ we get,

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

Now differentiating with respect to $x$

$⇒ \dfrac{x\left[2x + 2y \cdot \dfrac{dy}{dx}\right] - \left[{x}^{2} + {y}^{2} - 4\right] \cdot 1}{{x}^{2}} = 0$

$⇒ 2xy \cdot \dfrac{dy}{dx} + \left({x}^{2} - {y}^{2} + 4\right) = 0$