Question

Find the differential equation of all circle passing through the origin and having their centres on the x- axis is ?

Answer 1

Let, the equation of the circle assuming $R$ as radius and $(a,b)$ as the centre of the circle is,

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

Since, the radius of the circle passes through the origin with the centre on $x$-axis, so the centre of circle will be $(a,0)$.

Thus, equation of the circle is,

${(x-a)}^2+y^2=a^2⟶\left(1\right)$

Differentiating both sides we get,

$2(x-a)dx+2ydy=0\Rightarrow x-a=-\frac{ydy}{dx}anda=x+\frac{ydy}{dx}$

Substituting these values in equation $\left(1\right)$,

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

Hence, this is the required differential equation of all the circles passing through the origin and having their centres on the $x$-axis.