Question

Find the polar equation of a circle, the initial line being a tangent. What does it become if the origin be on the circumference?

Answer 1

The circle is shown in the figure, equation of the circle will be

$(x-a\cot \alpha {)}^2+(y-a{)}^2=a^2$

Converting it to polar form

$(r\cos \theta -a\cot \alpha {)}^2+(r\sin \theta -a{)}^2=a^2$

$r^2{\cos }^2\theta +a^2{\cot }^2\alpha -2ar\cos \theta \cot \alpha +r^2{\sin }^2\theta +a^2-2ar\sin \theta =a^2$

$r^2+a^2{\cot }^2\alpha -2ar\cos \theta \cos \alpha cosec\alpha -2ar\sin \theta \sin \alpha cosec\alpha =0$

$r^2+a^2{\cot }^2\alpha -2arcodec\alpha .\cos (\theta -\alpha )=0$

If the origin is on circumference, put $a\cot \alpha =0$

We get, $\Rightarrow$ the equation becomes $r=2a\sin \theta$