SATHEE: Complex Numbers 2 Question 37

Complex Numbers 2 Question 37

38.

Find the centre and radius of the circle formed by all the points represented by $z = x + i y$ satisfying the relation $| \frac{z - α}{z - β} | = k (k \neq 1)$ , where $α$ and $β$ are the constant complex numbers given by $α = α_{1} + i α_{2}, β = β_{1} + i β_{2}$

(2004, 2M)

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Answer:

Correct Answer: 38. Centre $= \frac{α - k^{2} β}{1 - k^{2}}$ , Radius $= | \frac{k (α - β)}{1 - k^{2}} |$

Solution:

As we know, $| z |^{2} = z \cdot \bar{z}$

Given, $\frac{| z - α |^{2}}{| z - β |^{2}} = k^{2}$

$\Rightarrow (z - α) (\bar{z} - \bar{α}) = k^{2} (z - β) (\bar{z} - \bar{β})$

$\Rightarrow | z |^{2} - α \bar{z} - \bar{α} z + | α |^{2} = k^{2} (| z |^{2} - β \bar{z} - \bar{β} z + | β |^{2})$

$\Rightarrow | z |^{2} (1 - k^{2}) - (α - k^{2} β) \bar{z} - (\bar{α} - \bar{β} k^{2}) z$

$\begin{aligned} + (| α |^{2} - k^{2} | β |^{2}) = 0 \\ \Rightarrow | z |^{2} - \frac{(α - k^{2} β)}{(1 - k^{2})} \bar{z} - \frac{(\bar{α} - \bar{β} k^{2})}{(1 - k^{2})} z + \frac{| α |^{2} - k^{2} | β |^{2}}{(1 - k^{2})} = 0 . \end{aligned}$

On comparing with equation of circle,

$| z |^{2} + a \bar{z} + \bar{a} z + b = 0$

whose centre is $(- a)$ and radius $= \sqrt{| a |^{2} - b}$

$∴$ Centre for Eq. (i) $= \frac{α - k^{2} β}{1 - k^{2}}$

$\begin{aligned} and radius & = \sqrt{\frac{α - k^{2} β}{1 - k^{2}}} \frac{\bar{α} - k^{2} \bar{β}}{1 - k^{2}} - \frac{α \bar{α} - k^{2} β \bar{β}}{1 - k^{2}} \\ = \frac{k (α - β)}{1 - k^{2}} \end{aligned}$

Complex Numbers 2 Question 37

38.

Answer:

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Complex Numbers 2 Question 37

38.

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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