SATHEE: Circle 3 Question 10

Circle 3 Question 10

10. Let $R S$ 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 $R S$ at point $E$ . Then, the locus of $E$ passes through the point(s)

(a) $\frac{1}{3}, \frac{1}{\sqrt{3}}$

(b) $\frac{1}{4}, \frac{1}{2}$

(d) $\frac{1}{4}, - \frac{1}{2}$

(2016 Adv.)

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Answer:

Correct Answer: 10. $(a, c)$

Solution:

Given, $R S$ is the diameter of $x^{2} + y^{2} = 1$ .

Here, equation of the tangent at $P (\cos θ, \sin θ)$ is $x \cos θ + y \sin θ = 1$ .

Intersecting with $x = 1$ ,

$\begin{aligned} y = \frac{1 - \cos θ}{\sin θ} \\ ∴ & Q 1, \frac{1 - \cos θ}{\sin θ} \end{aligned}$

$∴$ Equation of the line through $Q$ parallel to $R S$ is

$y = \frac{1 - \cos θ}{\sin θ} = \frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}} = \tan \frac{θ}{2} .$

Normal at $P : y = \frac{\sin θ}{\cos θ} \cdot x$

$\Rightarrow y = x \tan θ$

Let their point of intersection be $(h, k)$ .

$\begin{aligned} Then, k = \tan \frac{θ}{2} and k = h \tan θ \\ ∴ k = h \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}} \Rightarrow k = \frac{2 h \cdot k}{1 - k^{2}} \end{aligned}$

$\Rightarrow k (1 - k^{2}) = 2 h k$

$∴$ Locus for point $E : 2 x = (1 - y^{2})$

When $x = \frac{1}{3}$ , then

$1 - y^{2} = \frac{2}{3} \Rightarrow y^{2} = 1 - \frac{2}{3} \Rightarrow y = \pm \frac{1}{\sqrt{3}}$

$∴ \frac{1}{3}, \pm \frac{1}{\sqrt{3}}$ satisfy $2 x = 1 - y^{2}$ .

When $x = \frac{1}{4}$ , then

$1 - y^{2} = \frac{2}{4} \Rightarrow y^{2} = 1 - \frac{1}{2} \Rightarrow y = \pm \frac{1}{\sqrt{2}}$

$∴ \frac{1}{4}, \pm \frac{1}{2}$ does not satisfy $1 - y^{2} = 2 x$ .

Circle 3 Question 10

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Circle 3 Question 10

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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