The differential equation of family of circles, whose centres are on the X -axis and also touch the Y -axis is

The differential equation of family of circles, whose centres are on the X -axis and also touch the Y -axis is
  1. $4\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$
  2. $\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$
  3. $2\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$
  4. $\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=4\left(x^2+y^2\right)^2$

Solution

The system of circles whose centre lies on X -axis and touch Y -axis (i.e., passes through the origin) is $x^2+y^2=2 \mathrm{~b} x...(i)$
Differentiating w.r.t $x$, we get $x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\mathrm{b}...(ii)$
Substituting (ii) in (i), we get $\begin{aligned} & x^2+y^2=2\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right) x \\ & \Rightarrow\left(x^2+y^2\right)^2=4\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2 \end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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