Question

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

Answer 1

General equation of the circle is

${(x-a)}^2+{(y-b)}^2=r^2$

Given$:$Centre is on the $y-$ axis and radius $=3$ units

$\therefore$ centre$=(0,b)$

Hence, our equation. becomes

${(x-0)}^2+{(y-b)}^2=3^2$

$\Rightarrow x^2+{(y-b)}^2=3^2$ .......$\left(1\right)$

Differentiating both sides w.r.t $x$

$2x+2(y-b)[\frac{dy}{dx}-0]=0$

$\Rightarrow x+(y-b)\frac{dy}{dx}=0$

$\Rightarrow (y-b)\frac{dy}{dx}=-x$

$y-b=\frac{-x}{\frac{dy}{dx}}$

Put the value of $(y-b)$ in $\left(1\right)$ we get

$x^2+{(y-b)}^2=9$

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

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

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

$\Rightarrow x^2{y^{\prime }}^2+x^2=9{y^{\prime }}^2$

$\Rightarrow (x^2-9){y^{\prime }}^2+x^2=0$ is the required equation.