Question

Let $RS$ be the diameter of the circle $x^2+y^2=1$, where $S$ is the point $(1,0)$. Let $P$ be a variable point (other than $R$ and $S$) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$. The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $RS$ at point $E$. Then the locus of $E$ passes through the point(s).

• A. $(\frac{1}{3},\frac{1}{\sqrt{3}})$
• B. $(\frac{1}{4},\frac{1}{2})$
  
• C. $(\frac{1}{3},-\frac{1}{\sqrt{3}})$
  
• D. $(\frac{1}{4},-\frac{1}{2})$

Answer 1

Answer: (A), (C)

The correct options are
A $(\frac{1}{3},\frac{1}{\sqrt{3}})$
C $(\frac{1}{3},-\frac{1}{\sqrt{3}})$



Given equation of circle $x^2+y^2=1$
Let $P(\cos \theta ,\sin \theta )$ be any point on the circle
Tangent at $P$ is
$x\cos \theta +y\sin \theta -1=0$
Equation of line $QS:x=1$
Finding the point of intersection of line $QS$ and the tangent, we get
$\cos \theta +y\sin \theta -1=0\Rightarrow y=\frac{1-\cos \theta }{\sin \theta }\Rightarrow Q=(1,\frac{1-\cos \theta }{\sin \theta })$

Equation of normal at point $P$ is
$y\cos \theta -x\sin \theta +\lambda =0$
Putting point $P$, we get
$\lambda =0$
Therefore, the equation of normal is
$y=\left(\tan \theta \right)x$
Line $EQ:y=\left(\frac{1-\cos \theta }{\sin \theta }\right)x$

So, the point of intersection of normal and line $EQ$,
$E=(\frac{1-\cos \theta }{\sin \theta \tan \theta },\frac{1-\cos \theta }{\sin \theta })\Rightarrow E(\frac{\tan \frac{\theta }{2}}{\tan \theta },\tan \frac{\theta }{2})$

Assuming the point $E=(h,k)$, we get
$h=\frac{\tan \frac{\theta }{2}}{\tan \theta },k=\tan \frac{\theta }{2}$
Now,
$\Rightarrow \frac{k}{h}=\tan \theta =\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}\Rightarrow \frac{k}{h}=\frac{2k}{1-k^2}\Rightarrow k^3+2hk-k=0\Rightarrow k(k^2+2h-1)=0\Rightarrow k^2+2h-1=0$
Therefore, the locus of the point $E$ is
$2x+y^2-1=0$

By observation, we see that the points
$(\frac{1}{3},\frac{1}{\sqrt{3}})$ and $(\frac{1}{3},-\frac{1}{\sqrt{3}})$ satisfies the locus.