Question

Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.

Answer 1

The equation of the family of circles with radius 3 units, having its centre on y-axis, is given by

\[x^2 + \left( y - a \right)^2 = 3^2 . . . . . \left( 1 \right)\]

Here, a is any arbitrary constant.

Since this equation has only one arbitrary constant, we get a first order differential equation.

Differentiating (1) with respect to x, we get

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

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

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

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

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

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

\[x^2 + \left( y - \left(y + \frac{x}{\frac{dy}{dx}}\right) \right)^2 = 3^2 \]

\[x^2 + \left( y - y - \frac{x}{\frac{dy}{dx}} \right)^2 = 3^2 \]
\[ \Rightarrow x^2 + \frac{x^2}{\left( \frac{dy}{dx} \right)^2} = 9\]

\[ \Rightarrow x^2 \left( \frac{dy}{dx} \right)^2 + x^2 = 9 \left( \frac{dy}{dx} \right)^2 \]

\[ \Rightarrow x^2 \left( \frac{dy}{dx} \right)^2 - 9 \left( \frac{dy}{dx} \right)^2 + x^2 = 0\]

\[ \Rightarrow \left( x^2 - 9 \right) \left( \frac{dy}{dx} \right)^2 + x^2 = 0\]

\[ \Rightarrow \left( x^2 - 9 \right) \left( y' \right)^2 + x^2 = 0\]

\[\text{Hence, }\left( x^2 - 9 \right) \left( y' \right)^2 + x^2 = 0\text{ is the required differential equation.}\]