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
$$\begin{gathered} 2x+2\left(y-a\right)\frac{dy}{dx}=0 \\ \Rightarrow x+\left(y-a\right)\frac{dy}{dx}=0 \\ \Rightarrow x=\left(a-y\right)\frac{dy}{dx} \\ \Rightarrow \frac{x}{{\displaystyle \frac{dy}{dx}}}=a-y \\ \Rightarrow a=y+\frac{x}{{\displaystyle \frac{dy}{dx}}} \end{gathered}$$
Substituting the value of a in (1), we get

$$\begin{gathered} x^2+{\left(y-y-\frac{x}{{\displaystyle \frac{dy}{dx}}}\right)}^2=3^2 \\ \Rightarrow x^2+\frac{x^2}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2}=9 \\ \Rightarrow x^2{\left(\frac{dy}{dx}\right)}^2+x^2=9{\left(\frac{dy}{dx}\right)}^2 \\ \Rightarrow x^2{\left(\frac{dy}{dx}\right)}^2-9{\left(\frac{dy}{dx}\right)}^2+x^2=0 \\ \Rightarrow \left(x^2-9\right){\left(\frac{dy}{dx}\right)}^2+x^2=0 \\ \Rightarrow \left(x^2-9\right){\left(y\text{'}\right)}^2+x^2=0 \\ \mathrm{Hence},\left(x^2-9\right){\left(y\text{'}\right)}^2+x^2=0\text{is the required differential equation.} \end{gathered}$$