JEE Advanced · Mathematics · 12. Circle

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 $R & S$ ) on the circle and tangents to the circle at $S & P$ meet at the point $Q$ . The normal to the circle at P intersects a line drawn through $Q$ parallel to $R S$ 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 $(\frac{1}{4}, - \frac{1}{2})$

Verified Solution

Answer & Solution

Correct Answer

(C) $(\frac{1}{3}, - \frac{1}{\sqrt{3}})$

Step-by-step Solution

Detailed explanation

Let P be (cos θ, sin θ)

, where

θ \neq 0, π

Tangent at

P : x \cos θ + y \sin θ = 1

........(i)
Tangent at

S : x = 1

.........(ii)

∴

By (i) and (ii) :

Q (1, \frac{1 - \cos θ}{\sin θ})

Line through Q parallel to RS :

y = \frac{1 - \cos θ}{\sin θ} \Rightarrow y = \tan \frac{θ}{2}

..........(iii)
Normal at

P : y = \frac{\sin θ}{\cos θ} x \Rightarrow y = \tan θ . x

\Rightarrow y = (\frac{2 t an \frac{θ}{2}}{1 - t {an}^{2} \frac{θ}{2}}) x

........(iv)
Point of intersection of equation (iii) and (iv),

E : h = \frac{1 - \tan^{2} \frac{θ}{2}}{2}; k = \tan \frac{θ}{2}

Eliminating

θ : h = \frac{1 - k^{2}}{2} \Rightarrow y^{2} = 1 - 2 x

(\frac{1}{3}, \frac{1}{\sqrt{3}}) & (\frac{1}{3}, - \frac{1}{\sqrt{3}})

satisfy the locus.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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