Question

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

Answer 1

The equation of the family of circles that pass through the origin (0,0) and whose centres lie on the x-axis is given by
${\left(x-a\right)}^2+y^2=a^2$ ...(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
$$\begin{gathered} 2\left(x-a\right)+2y\frac{dy}{dx}=0 \\ \Rightarrow x-a+y\frac{dy}{dx}=0 \\ \Rightarrow x+y\frac{dy}{dx}=a..\text{(2)} \end{gathered}$$
Substituting the value of $a$ in equation (1), we get
$$\begin{gathered} {\left(x-x-y\frac{dy}{dx}\right)}^2+y^2={\left(x+y\frac{dy}{dx}\right)}^2 \\ \Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2+y^2=x^2+2xy\frac{dy}{dx}+y^2{\left(\frac{dy}{dx}\right)}^2 \\ \Rightarrow 2xy\frac{dy}{dx}+x^2=y^2 \\ \mathrm{It}\mathrm{is}\mathrm{the}\text{required differential equation.} \end{gathered}$$