The differential equation of all circles which pass through the origin and whose centres lie on $\mathrm{Y}$-axis is

Question

Aicharya · Accepted Answer

Circle passes through origin and centre lie on $\mathrm{Y}$-axis. Let $(0, k)$ be centre and ' $k$ ' be radius $	herefore \quad$ Equation of circle is $\begin{aligned} & (x-0)^2+(y-\mathrm{k})^2=\mathrm{k}^2 \ & x^2+y^2-2 y \mathrm{k}+\mathrm{k}^2=\mathrm{k}^2 \ & x^2+y^2-2 \mathrm{k} y=0 \ & x^2+y^2=2 \mathrm{k} y ...(i)\ & \frac{x^2+y^2}{2 y}=\mathrm{k} ...(ii) \end{aligned}$ Differentiating equation (i) with respect to $x$, we get $\begin{aligned} & 2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{k} \frac{\mathrm{d} y}{\mathrm{~d} x} \ & 2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}-2 \mathrm{k} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \ & 2 x+2(y-\mathrm{k}) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \ & 2 x+2\left[y-\left(\frac{x^2+y^2}{2 y}ight)ight] \frac{\mathrm{d} y}{\mathrm{~d} x}=0 ...[From(ii)]\ & 2 x+2\left[\frac{2 y^2-x^2-y^2}{2 y}ight] \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \ & 2 x+\left(\frac{y^2-x^2}{y}ight) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \ & 2 x y+\left(y^2-x^2ight) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \ & 	ext { i.e. }\left(x^2-y^2ight) \frac{\mathrm{d} y}{\mathrm{~d} x}-2 x y=0 \end{aligned}$

The differential equation of all circles which pass through the origin and whose centres lie on…

Solution