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

\[2x + 2\left( y - a \right)\frac{dy}{dx} = 0\]

\[ \Rightarrow x + \left( y - a \right)\frac{dy}{dx} = 0\]

\[ \Rightarrow x = \left( a - y \right)\frac{dy}{dx}\]

\[ \Rightarrow \frac{x}{\frac{dy}{dx}} = a - y\]

\[ \Rightarrow a = y + \frac{x}{\frac{dy}{dx}} ................(2)\]

Substituting the value of a in equation (2), we get

\[x^2 + \left( y - y - \frac{x}{\frac{dy}{dx}} \right)^2 = \left( y + \frac{x}{\frac{dy}{dx}} \right)^2 \]

\[ \Rightarrow x^2 + \frac{x^2}{\left( \frac{dy}{dx} \right)^2} = y^2 + 2\frac{xy}{\frac{dy}{dx}} + \frac{x^2}{\left( \frac{dy}{dx} \right)^2}\]

\[ \Rightarrow x^2 = y^2 + 2\frac{xy}{\frac{dy}{dx}}\]

\[ \Rightarrow \left( x^2 - y^2 \right)\frac{dy}{dx} = 2xy\]

It is the required differential equation.