Question

Find the polar equation to the circle described on the straight line joining the points $(a,\alpha )$ and $(b,\beta )$ as diameter.

Answer 1

Equation of circle with endpoints of diameter $(x_1,y_1),(x_2,y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

putting $(x,y)=(r\cos \theta ,r\sin \theta )$

The endpoints of diameter in polar form are $(a\cos \alpha ,a\sin \alpha )\&\left(b\cos \beta ,b\sin \beta \right)$

$\therefore$ the equation becomes $(rcos\theta -acos\alpha )(rcos\theta -bcos\beta )+(rsin\theta -asin\alpha )(rsin\theta -bsin\beta )=0$

$r^{2}co{s}^2\theta -arcos\alpha cos\theta -brcos\theta cos\beta +abcos\alpha cos\beta +r^{2}si{n}^2\theta -arsin\alpha sin\theta +absin\alpha sin\beta -rbsin\beta sin\theta =0$

On solving, we get $r^2-arcos(\theta -\alpha )-brcos(\theta -\beta )+abcos(\alpha -\beta )=0$