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

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

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

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

Substituting the value of a in equation (1), we get

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

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

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

It is the required differential equation.