Question

Find the differential equation of all the circles which pass through the origin and whose centres lie on y-axis.

Answer 1

The equation of the family of circles that pass through the origin (0, 0) and whose centres lie on the y-axis is given by
$x^2+{\left(y-a\right)}^2=a^2$ ...(1)
where $a$ is any arbitrary constant.
As this equation has only one arbitrary constant, we shall get a first order differential equation.
Differentiating equation (1) with respect to $x$, we get
$$\begin{gathered} 2x+2\left(y-a\right)\frac{dy}{dx}=0 \\ \Rightarrow x+\left(y-a\right)\frac{dy}{dx}=0 \\ \Rightarrow x=\left(a-y\right)\frac{dy}{dx} \\ \Rightarrow \frac{x}{{\displaystyle \frac{dy}{dx}}}=a-y \\ \Rightarrow a=y+\frac{x}{{\displaystyle \frac{dy}{dx}}}...\text{(2)} \end{gathered}$$
Substituting the value of $a$ in equation (2), we get
$$\begin{gathered} x^2+{\left(y-y-\frac{x}{{\displaystyle \frac{dy}{dx}}}\right)}^2={\left(y+\frac{x}{{\displaystyle \frac{dy}{dx}}}\right)}^2 \\ \Rightarrow x^2+\frac{x^2}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2}=y^2+2\frac{xy}{{\displaystyle \frac{dy}{dx}}}+\frac{x^2}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2} \\ \Rightarrow x^2=y^2+2\frac{xy}{{\displaystyle \frac{dy}{dx}}} \\ \Rightarrow \left(x^2-y^2\right)\frac{dy}{dx}=2xy \\ \mathrm{It}\mathrm{is}\mathrm{th}\text{e required differential equation.} \end{gathered}$$