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 $\left|\frac{z-\alpha}{z-\beta}\right|=k(k \neq 1)$, where $\alpha$ and $\beta$ are the constant complex numbers given by $\alpha=\alpha _1+i \alpha _2, \beta=\beta _1+i \beta _2$
(2004, 2M)
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Answer:
Correct Answer: 38. Centre $=\frac{\alpha-k^{2} \beta}{1-k^{2}}$, Radius $=\left|\frac{k(\alpha-\beta)}{1-k^{2}}\right|$
Solution:
- As we know, $|z|^{2}=z \cdot \bar{z}$
Given, $\quad \frac{|z-\alpha|^{2}}{|z-\beta|^{2}}=k^{2}$
$\Rightarrow \quad(z-\alpha)(\bar{z}-\bar{\alpha})=k^{2}(z-\beta)(\bar{z}-\bar{\beta})$
$\Rightarrow|z|^{2}-\alpha \bar{z}-\bar{\alpha} z+|\alpha|^{2}=k^{2}\left(|z|^{2}-\beta \bar{z}-\bar{\beta} z+|\beta|^{2}\right)$
$\Rightarrow|z|^{2}\left(1-k^{2}\right)-\left(\alpha-k^{2} \beta\right) \bar{z}-\left(\bar{\alpha}-\bar{\beta} k^{2}\right) z$
$ \begin{aligned} & +\left(|\alpha|^{2}-k^{2}|\beta|^{2}\right)=0 \\ & \Rightarrow|z|^{2}-\frac{\left(\alpha-k^{2} \beta\right)}{\left(1-k^{2}\right)} \bar{z}-\frac{\left(\bar{\alpha}-\bar{\beta} k^{2}\right)}{\left(1-k^{2}\right)} z+\frac{|\alpha|^{2}-k^{2}|\beta|^{2}}{\left(1-k^{2}\right)}=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 $=\sqrt{|a|^{2}-b}$
$\therefore$ Centre for Eq. (i) $=\frac{\alpha-k^{2} \beta}{1-k^{2}}$
$ \begin{aligned} \text { and radius } & =\sqrt{\frac{\alpha-k^{2} \beta}{1-k^{2}}} \frac{\bar{\alpha}-k^{2} \bar{\beta}}{1-k^{2}}-\frac{\alpha \bar{\alpha}-k^{2} \beta \bar{\beta}}{1-k^{2}} \\ & =\frac{k(\alpha-\beta)}{1-k^{2}} \end{aligned} $
