Question

Find the center, radius of the circle given by $\left|\frac{(z-\alpha )}{(z-\beta )}\right|=k,k\ne 1$, where $z=x+iy,a=a_1+i{a}_2,\beta ={\beta }_1+i{\beta }_2$.

• A. Center = $\frac{\alpha +k^2\beta }{1-k^2}$ and $radius=\frac{k|\alpha -\beta |}{|1-k^2|}$
• B. Center = $\frac{\alpha -k^2\beta }{1-k^2}$ and $radius=\frac{k|\alpha -\beta |}{|1-k^2|}$
• C. Center = $\frac{\alpha -k^2\beta }{1+k^2}$ and $radius=\frac{k|\alpha -\beta |}{|1-k^2|}$
• D. Center = $\frac{\alpha -k^2\beta }{1-k^2}$ and $radius=\frac{k|\alpha +\beta |}{|1-k^2|}$

Answer 1

The correct option is B Center = $\frac{\alpha -k^2\beta }{1-k^2}$ and $radius=\frac{k|\alpha -\beta |}{|1-k^2|}$

As we know ${\left|z\right|}^2=z.\overline{z}$
Given $\frac{{|z-\alpha |}^2}{{|z-\beta |}^2}=k^2$
$\Rightarrow (z-\alpha )(\overline{z}-\overline{\alpha })=k^2(z-\beta )(\overline{\alpha }-\overline{\beta })\Rightarrow {\left|z\right|}^2-\alpha \overline{z}-\overline{\alpha }z+{\left|\alpha \right|}^2=k^2({\left|z\right|}^2-\beta \overline{z}-\overline{\beta z}+{\left|\beta \right|}^2)$
$\Rightarrow {\left|z\right|}^2-\frac{(\alpha -k^2\beta )}{(1-k^2)}\overline{z}-\frac{(\overline{\alpha }-\overline{\beta }{k}^2)}{(1-k^2)}z+\frac{{\left|\alpha \right|}^2-k^2{\left|\beta \right|}^2}{(1-k^2)}=0$ ...(1)
On comparing with equation of circle
${\left|z\right|}^2+\alpha \overline{z}+\overline{\alpha }z+\beta =0$
whose center is $(-\alpha )$ and radius $=\sqrt{{\left|\alpha \right|}^2-b}$
Therefore center for equation (1) $=\frac{\alpha -k^2\beta }{1-k^2}$
and radius $=\sqrt{\left(\frac{\alpha -k^2\beta }{1-k^2}\right)\left(\frac{\overline{\alpha }-k^2\overline{\beta }}{1-k^2}\right)-\frac{\alpha \overline{\alpha }-k^2\beta \overline{\beta }}{1-k^2}}=\left|\frac{k(\alpha -\beta )}{1-k^2}\right|$