Question

If $(2,4)$ is a point on the orthogonal trajectory of $x^2+y^2-ay=0$ then the orthogonal trajectory is a circle with radius

Answer 1

Differentiating w.r.t. $x$, we have
$2x+2y\frac{dy}{dx}-a\frac{dy}{dx}=0.\Rightarrow a=2(x\frac{dx}{dy}+y)$.
Substituting this value in the given equation, we have
$x^2+y^2-2(x\frac{dx}{dy}+y)y=0\Rightarrow x^2-y^2-2xy\frac{dx}{dy}=0$.
Replacing $\frac{dy}{dx}by-\frac{dx}{dy}$, we have
$x^2-y^2+2xy\frac{dy}{dx}=0.\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
This is a homogeneous equation in $x$ and $y$.
Substituting $y=Vx$, we have
$V+x\frac{dV}{dx}=\frac{V^2{x}^2-x^2}{2x.Vx}=\frac{V^2-1}{2V}$
$x\frac{dV}{dx}=\frac{V^2-1-2V^2}{2V}=-\frac{1+V^2}{2V}$
$\Rightarrow \frac{dx}{x}=\frac{-2V}{1+V^2}dV$
$\Rightarrow \log x+\log (1+V^2)=const$
$\Rightarrow x(1+y^2/x^2)=const=b$
$\Rightarrow x^2+y^2-bx=0$ is required orthogonal trajectories.
Since $(2,4)$ lies on this curve so $b=10$. Thus $x^2+y^2-10x=0$ which is circle of radius $5$