Question

Obtain the differential equation of the family of circle passing through the points $(a,0)$ and $(-a,0)$?

Answer 1

Since the circle pass through the points $(a,0)$ and $(a,0)$ , the circle of the circle will be the circle will have radius $=a$

So, the general equation of such a circle is

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

The equation should be difference once since there is only one arbitrary constant , $a$

Differentiating (i) on both sides, we get

$\begin{array}{c}2\left(x-a\right)+2\frac{dy}{dx}=............\left(ii\right)\\ \Rightarrow x-a=-y\frac{dy}{dx}anda=x+y\frac{dy}{dx}\end{array}$

Substituting the values of $x-a$ and $a$ in (i), we get

$\begin{array}{c}{\left(y\frac{dy}{dx}\right)}^2+y^2={\left(x+y\frac{dy}{dx}\right)}^2\\ \Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2+y^2=x^2+2xy\frac{dy}{dx}+y^2{\left(\frac{dy}{dx}\right)}^2\\ \Rightarrow y^2=x^2+2xy\frac{dy}{dx}\end{array}$

Which is the required differential equation.