The differential equation of the family of circles passing through the origin and having c…

Question

The differential equation of the family of circles passing through the origin and having centre at the line $y=x$ is :

Accepted Answer

Equation of circle passing through origin \& having centre at the line $y=x$ is

$$\begin{aligned}
& (x-t)^2+(y-t)^2=2 t^2 \
& x^2+y^2+t^2+t^2-2 t x-2 t y=2 t^2 \
& x^2+y^2=2 t(x+y)
\end{aligned}$$

Now differentiate

$$\begin{aligned}
& 2 x+2 y y^{\prime}=2 t\left(1+y^{\prime}\right) \
& t=\frac{x+y y^{\prime}}{1+y^{\prime}}
\end{aligned}$$

Now, $$x^2+y^2=2\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)(x+y)$$

$$\begin{aligned}
& x^2+y^2+x^2 \frac{d y}{d x}+y^2 \frac{d y}{d x} \
& =2\left(x^2+x y+x y \frac{d y}{d x}+y^2 \frac{d y}{d x}\right) \
& d x\left(x^2+y^2-2 x^2-2 x y\right)=d y\left(2 x y+2 y^2-x^2-y^2\right) \
& d x\left(x^2-y^2+2 x y\right)=d y\left(x^2-y^2-2 x y\right)
\end{aligned}$$

Solutions · 1