Question

If $\left(\frac{3-z_1}{2-z_1}\right)\left(\frac{2-z_2}{3-z_2}\right)=k;$ then points $A\left(z1\right),B\left(z2\right),C(2,0)$ and $D(3,0)$ will

• A. lie on a circle only for k > 0
• B. lie on a circle only for k < 0
• C. lie on a circle R
• D. be vertices of a square ( 0, 1)

Answer 1

The correct option is A lie on a circle only for k > 0

Arg$\left(\frac{3-z_1}{2-z_1}\right)+arg\left(\frac{2-z_2}{3-z_2}\right)$

=arg$\left(\frac{3-z_1}{2-z_1}\right)\left(\frac{2-z_2}{3-z_2}\right)$

Now if $\left(\frac{3-z_1}{2-z_1}\right)\left(\frac{2-z_2}{3-z_2}\right)$ is a +ve real

number, then its argument will be zero

So, angles ${\theta }_1$and ${\theta }_2$ are equal in magnitude but opposite in sign. So chord DC subtends equal angles at A and B. So points are concyclic for $k>0$