If $|\frac{(z - z_{1})}{(z - z_{2})}| = 3$ , where $z_{1} a n d z_{2}$ are fixed complex numbers and $z$ is a variable complex number, then $z$ lies on a

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circle with $z_{1}$ as its interior point

circle with $z_{2}$ as its interior point

circle with $z_{1}$ as its exterior point

circle with $z_{2}$ as its exterior point

answer is B, C.

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Detailed Solution

If $z_{1}, z_{2} a n d z$ represent the points $A, B a n d P$ on the argand plane, then
$\frac{P A}{P B} = \frac{3}{1} \Rightarrow A P : P B = 3 : 1$
Let the internal and external angular bisectors of $∠ A P B$ meet the line joining A and B at $P_{1} a n d P_{2}$ , respectively. Hence
$A P_{1} : P_{1} B = A P : P B = 3 : 1$ , internally
$A P_{2} : P_{2} B = A P : P B = 3 : 1$ , externally
Thus $P_{1} a n d P_{2}$ are fixed points. Also, $∠ P_{1} P P_{2} = \frac{π}{2}$ .
Thus P lies on a circle having $P_{1} P_{2}$ as its diameter. Clearly $B (z_{2})$ lies inside this circle.