Question

The locus of foot of the perpendicular from pole to the tangent to the circle $r=2a\cos \theta$ is

• A. $r=a(1+\cos \theta$)
• B. $r=a(1-\cos \theta$)
• C. $r=a(1+\sin \theta$)
• D. $r=a(1-\sin \theta$)

Answer 1

The correct option is A $r=a(1+\cos \theta$)

Refer to the figure,
OC is the perpendicular drawn to the tangent from the pole. $C$ is the foot of the perpendicular. Let $OC=R$ and $OC$ makes an angle $\alpha$ with the initial line $OB$.
$B$ is the center of the circle. Equation of the circle is $r=2a\cos \theta$
So, $OB=AB=a$
$BA\perp AC$ (since AC is the tangent)
$\angle AOB=\angle OAB=\theta$
In triangle OAC, $\angle COA=\theta$ (since right angled triangle)
So, we get $\alpha =2\theta$
In triangle OAC, $R=r\cos \theta$
$\Rightarrow R=2a\cos \theta .\cos \theta$
$\Rightarrow R=a(1+\cos 2\theta )$
$\Rightarrow R=a(1+\cos \alpha )$
which is the relation between $R$ and $\alpha$ of the foot of the perpendicular; hence the locus of the foot of the perpendicular.