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Question: If P and Q are represented by the complex numbers z<sub>1</sub> and z<sub>2</sub>, such that \(\left...

If P and Q are represented by the complex numbers z1 and z2, such that ∣1z2+1z1∣\left| \frac{1}{z_{2}} + \frac{1}{z_{1}} \right|= ∣1z2−1z1∣\left| \frac{1}{z_{2}} - \frac{1}{z_{1}} \right|, then the circumcentre of

DOPQ (where O is the origin) is

A

12\frac{1}{2} (z1 – z2)

B

13\frac{1}{3} (z1 + z2)

C

12\frac{1}{2} (z1 + z2)

D

13\frac{1}{3} (z1 – z2)

Answer

12\frac{1}{2} (z1 + z2)

Explanation

Solution

Sol. Since∣1z2+1z1∣\left| \frac{1}{z_{2}} + \frac{1}{z_{1}} \right|= ∣1z2–1z1∣\left| \frac{1}{z_{2}}–\frac{1}{z_{1}} \right|Ž | z1 + z2 | = | z1 – z2 |

Squaring both sides, we have | z1 |2 + | z2 |2 + 2(z1zˉ2+zˉ1z2)2(z_{1}{\bar{z}}_{2} + {\bar{z}}_{1}z_{2})

= | z1 |2 + | z2 |2 – 2(z1zˉ2+zˉ1z2)2(z_{1}{\bar{z}}_{2} + {\bar{z}}_{1}z_{2})

Ž 4(z1zˉ2+zˉ1z2)4(z_{1}{\bar{z}}_{2} + {\bar{z}}_{1}z_{2}) = 0 Ž(z1z2)\left( \frac{z_{1}}{z_{2}} \right)=(z1‾z2)\left( \frac{\overline{z_{1}}}{z_{2}} \right)

\ z1z2\frac{z_{1}}{z_{2}}is purely imaginary

Ž arg(z1z2)\left( \frac{z_{1}}{z_{2}} \right) = π2\frac{\pi}{2} = arg(z1–0z2–0)\left( \frac{z_{1}–0}{z_{2}–0} \right)

i.e. angle between z2, O and z1 is a right angle, taken in order. As shown in the above arrangement. Now, the circumcentre of the above arrangement will lie on the line PQ as diameter and is represented by C which is the centre of PQ, such that

z = z1+z22\frac{z_{1} + z_{2}}{2}; where, z is the affix of circumcentre.