Question

Find the differential equation of the family of circles $(x-a{)}^2-(y-b{)}^2=r^2.$

Answer 1

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

Since, $r$ is a constant, we have two arbitrary constants $a$ and $b$.

Differentiating $\left(1\right)$ w.r.t $x$, we have

$2(x-a)+2(y-b)\frac{dy}{dx}=0$ .....$\left(2\right)$

Differentiating $\left(2\right)$ w.r.t $x$, we have

$1+(y-b)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$

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

From eqn$\left(2\right)$ we have $(x-a)=-(y-b)\frac{dy}{dx}$

Put this value in eqn$\left(1\right)$ we have

${(-(y-b)\frac{dy}{dx})}^2+{(y-b)}^2=r^2$

$[{\left(\frac{dy}{dx}\right)}^2+1]{(y-b)}^2=r^2$ ....$\left(4\right)$

Squaring $\left(3\right)$ we have

${(y-b)}^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2={[1+{\left(\frac{dy}{dx}\right)}^2]}^2$ ......$\left(5\right)$

Dividing eqn$\left(4\right)$ by $\left(5\right)$ we have

$\frac{[{\left(\frac{dy}{dx}\right)}^2+1]}{{\left(\frac{d^{2}y}{d{x}^2}\right)}^2}=\frac{r^2}{{[{\left(\frac{dy}{dx}\right)}^2+1]}^2}$

$\Rightarrow {[1+{\left(\frac{dy}{dx}\right)}^2]}^3=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^3$ is the required differential equation.