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Question: If ![](https://cdn.pureessence.tech/canvas_293.png?top_left_x=251&top_left_y=0&width=300&height=260)...

If (2z23z2)\left( \frac{2 - z_{2}}{3 - z_{2}} \right)= k, then points A(z1), B(z2), C(3, 0) and D(2, 0) taken clockwise –Q

A

Lie on a circle only if k > 0

B

Lie on a circle only if k < 0

C

Lie on a circle " k Ī R

D

Are vertices of a square " k Ī (0, 1)

Answer

Lie on a circle only if k > 0

Explanation

Solution

Sol. arg (3z12z1)\left( \frac{3 - z_{1}}{2 - z_{1}} \right)+ arg (2z23z2)\left( \frac{2 - z_{2}}{3 - z_{2}} \right)= arg (3z12z1.2z23z2)\left( \frac{3 - z_{1}}{2 - z_{1}}.\frac{2 - z_{2}}{3 - z_{2}} \right)

Now if k Ī R+ then q1 + q2 = 0

So q1 and q2 are equal in magnitude but opposite in sign.

So DC chord subtends equal angles at A and B.

So points are concyclic for k > 0.