Question

Shew that the equation to the circle, which passes through the focus and touches the curve $\frac{l}{r}=1-e\cos \theta$ at the point $\theta =\alpha ,$ is
$r(1-e\cos \alpha {)}^2=l\cos (\theta -\alpha )-el\cos (\theta -2\alpha ).$

Answer 1

Let the equation to the circle passing through the focus be $r=2R\cos (\theta -\Phi )$ .....(1)
so that its center will be $(R,\Phi )$
As the circle touches the given conic $r=\frac{l}{1-e\cos \theta }$
at the point where $\theta =\alpha$, we have
$\frac{l}{1-e\cos \theta }=2R\cos (\alpha -\Phi )$ ....(2)
Now as the center of circle lies on the normal at any point on it, hence the point $(R,\Phi )$ must lie on the normal $\theta =\alpha$
So, $\sin (\theta -\alpha )-e\sin \varphi =\frac{-e\sin \alpha }{1-e\cos \alpha }\frac{l}{R}$
$\Rightarrow \sin (\alpha -\varphi )+e\sin \varphi =\frac{e\sin \alpha }{1-e\cos \alpha }\frac{l}{R}$
$=\frac{e\sin \alpha }{1-e\cos \alpha }\frac{2R\cos (\alpha -\varphi )(1-e\cos \alpha )}{R}$
$=e.2\sin \alpha \cos (\alpha -\varphi )$
Hence, $\sin (\alpha -\varphi )+e\sin \varphi =e[\sin \varphi -\sin (\varphi -2\alpha )]$
$\Rightarrow \sin \alpha \cos \varphi -\cos \alpha \sin \varphi =e[\sin \varphi \cos 2\alpha -\cos \varphi \sin 2\alpha ]$
$\Rightarrow \cos \varphi (\sin \alpha -e\sin 2\alpha )=\sin \varphi (\cos \alpha -e\cos 2\alpha )$
$\Rightarrow \frac{\sin \varphi }{\sin \alpha -e\sin 2\alpha }=\frac{\cos \varphi }{\cos \alpha -e\cos 2\alpha }$ .....(3)
Eliminating R from $1$ and $2$ we get
Eliminating $\varphi$ from (4) by relation (3), we get
$l\{\cos \theta (\cos \alpha -e\cos 2\alpha )+\sin \theta (\sin \alpha -e\sin 2\alpha )\}=r{(1-e\cos \alpha )}^2$
$\Rightarrow l\cos (\theta -\alpha )-el\cos (\theta -2\alpha )=r{(1-e\cos \alpha )}^2$