Question

Write the order of the differential equation of the family of circles touching X-axis at the origin.

Answer 1

The equation of the family of circles touching x-axis at the origin is
$$\begin{gathered} {\left(x-0\right)}^2+{\left(y-a\right)}^2=a^2 \\ {x}^2+y^2-2ay=0.....\left(1\right) \\ \mathrm{Here},a\mathrm{is}\mathrm{the}\mathrm{parameter}. \\ \mathrm{Since}\mathrm{this}\mathrm{equation}\mathrm{contains}\mathrm{only}\mathrm{one}\mathrm{arbitary}\mathrm{constant},\mathrm{we}\mathrm{differentiate}\mathrm{it}\mathrm{only}\mathrm{once}. \\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ 2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0 \\ a=\frac{x+\mathrm{y}\left({\displaystyle \raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}\right)}{{\displaystyle \raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}}.....\left(2\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathit{}a\mathit{}\mathrm{from}\left(2\right)\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ {x}^2+y^2=2y\left\{\frac{x+y\left({\displaystyle \raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}\right)}{{\displaystyle \raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}}\right\} \\ \left(x^2-y^2\right)\frac{dy}{dx}=2xy \\ \mathrm{So},\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Here},\mathrm{order}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\mathrm{is}1. \end{gathered}$$