Question

If $\left|\frac{z-z_1}{z-z_2}\right|=3$, where $z_1$ and $z_2$ are fixed complex numbers and $z$ is a variable complex number, then '$z$' lies on a

• A. Circle with centre as $\frac{9z_1-z_2}{8}$
• B. Circle with ${}^{\prime }{z}_2^{\prime }$ as its interior point
• C. Circle with centre as $\frac{9z_2-z_1}{8}$
• D. Circle with ${}^{\prime }{z}_2^{\prime }$ as its exterior point

Answer 1

Answer: (B), (C)

The correct options are
B Circle with ${}^{\prime }{z}_2^{\prime }$ as its interior point
C Circle with centre as $\frac{9z_2-z_1}{8}$

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 $3:1$ 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,
$A{P}_1:P_{1}B\equiv PA:PB\equiv 3:1$ (internal division)
and
$A{P}_2:P_{2}B\equiv PA:PB\equiv 3:1$ (external division)
$\Rightarrow \angle P_{1}P{P}_2=\frac{\pi }{2}$
Thus locus of '$P$' is a circle having $P_1{P}_2$ as its diameter.
Now $P_1=\frac{z_1+3z_2}{4},P_2=\frac{3z_2-z_1}{2}$
Hence centre of circle will be $=\frac{P_1+P_2}{2}$
$=\frac{9z_2-z_1}{8}$