Let z1, z2 and z3 are the points on the argand plane which lie on the circle with equation |z - z0|=(4/√3) (where z0 is the centre of the circle). If z1=0, z2=-4 and z3=4+3z0, then (where arg Z ∈(-π, π]) )

Question

Let z1, z2 and z3 are the points on the argand plane which lie on the circle with equation |z - z0|=(4/√3) (where z0 is the centre of the circle). If z1=0, z2=-4 and z3=4+3z0, then (where arg Z ∈(-π, π]) ) (A) (π /6) (B) (5 π /12) (C) (5 π /6) (D) (2 π /3)

Tardigrade Admin · Accepted Answer

Here (z1 + z2 + z3/3)=z0⇒ z3=4+3z0 Therefore, center coincides with the circumcentre ⇒ Triangle is equilateral ⇒ |z1 - z2|=4 Clearly, z3 either lie in the second or third quadrant So the centre z0 also lies in the second or third quadrant. ∴ z0 can be =-2+(2/√3) texti,-2-(2/√3) texti ⇒ arg(z0)=(5 π /6),(7 π /6)

Q. Let $z_{1}, z_{2}$ and $z_{3}$ are the points on the argand plane which lie on the circle with equation $∣ z - z_{0} ∣ = \frac{4}{3}$ (where $z_{0}$ is the centre of the circle). If $z_{1} = 0, z_{2} = - 4$ and $z_{3} = 4 + 3 z_{0},$ then (where arg $Z \in (- π, π])$ )

$\frac{π}{6}$

$\frac{5 π}{12}$

$\frac{5 π}{6}$

$\frac{2 π}{3}$

Q. Let z1​,z2​ and z3​ are the points on the argand plane which lie on the circle with equation ∣z−z0​∣=3​4​ (where z0​ is the centre of the circle). If z1​=0,z2​=−4 and z3​=4+3z0​, then (where arg Z∈(−π,π]) )

6π​

125π​

65π​

32π​

Q. Let $z_{1}, z_{2}$ and $z_{3}$ are the points on the argand plane which lie on the circle with equation $∣ z - z_{0} ∣ = \frac{4}{3}$ (where $z_{0}$ is the centre of the circle). If $z_{1} = 0, z_{2} = - 4$ and $z_{3} = 4 + 3 z_{0},$ then (where arg $Z \in (- π, π])$ )

$\frac{π}{6}$

$\frac{5 π}{12}$

$\frac{5 π}{6}$

$\frac{2 π}{3}$