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Question: If \(\left( \frac{3 - z_{1}}{2 - z_{1}} \right)\left( \frac{2 - z_{2}}{3 - z_{2}} \right)\) = k, the...

If (3−z12−z1)(2−z23−z2)\left( \frac{3 - z_{1}}{2 - z_{1}} \right)\left( \frac{2 - z_{2}}{3 - z_{2}} \right) = k, then points A(z1), B(z2), C(3, 0) and D (2, 0) (taken in clockwise sense) will

A

Lie on a circle only for k > 0

B

Lie on a circle only for k < 0

C

Lie on a circle " k Ī R

D

Be the vertices of a square " k Ī (0, 1)

Answer

Lie on a circle only for k > 0

Explanation

Solution

Sol. arg (3−z12−z1)\left( \frac{3 - z_{1}}{2 - z_{1}} \right)+ arg (2−z23−z2)\left( \frac{2 - z_{2}}{3 - z_{2}} \right)

= arg (3−z12−z1.2−z23−z2)\left( \frac{3 - z_{1}}{2 - z_{1}}.\frac{2 - z_{2}}{3 - z_{2}} \right) = q1 + q2 = 0

Now if, (3−z12−z1.2−z23−z2)\left( \frac{3 - z_{1}}{2 - z_{1}}.\frac{2 - z_{2}}{3 - z_{2}} \right) is a positive real number, then its argument will be zero. So chord DC subtends equal angle at A and B. So points are concyclic for k > 0.