Question

Diameter of the circle given by $|(z-\alpha )/(z-\beta )|=k,k\ne 1$ , where $\alpha ,\beta$ are fixed points and $z$ is varying point in argand plane is

• A. $\frac{k|\alpha -\beta |}{|1-k^2|}$
• B. $\frac{2k|\alpha -\beta |}{|1-k^2|}$
• C. $\frac{3k|\alpha -\beta |}{|1-k^2|}$
• D. $\frac{4k|\alpha -\beta |}{|1-k^2|}$

Answer 1

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

Let $P_1,P_2$ be two points on the line joining the points $A\left(\alpha \right),B\left(\beta \right)$ which divides $A$ and $B$ in the ration $k:1$ internally and externally

Now internal and external bisectors of $\angle APB$ will meet the line joining points $A$ and $B$ at $P_1$ and $P_2$, respectively.
Since,
$A{P}_1:P_{1}B\equiv PA:PB\equiv k:1$ (internal division)
and
$A{P}_2:P_{2}B\equiv PA:PB\equiv k:1$ (external division)
$\Rightarrow \angle P_{1}P{P}_2=\frac{\pi }{2}$
Thus '$P$' lies on a circle having $P_1{P}_2$ as its diameter.
Now $P_1\left(z_1\right)=\frac{\alpha +k\left(\beta \right)}{k+1}$,
$P_2\left(z_2\right)=\frac{\alpha -k\left(\beta \right)}{1-k}$

So, diameter$=|P_1{P}_2|=|z_1-z_2|$
$=\frac{|(\alpha +k\beta )(1-k)-(\alpha -k\beta )(1+k)|}{|1-k^2|}=\frac{2k|\alpha -\beta |}{|1-k^2|}$