A point $z$ moves in the complex plane such that arg $\left(\frac{z-2}{z+2}\right)=\frac{\pi}{4}$, then the minimum value of $|z-9 \sqrt{2}-2 i|^{2}$ is equal to ____.

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Complex Numbers

A point

z

moves in the complex plane such that arg

(\frac{z - 2}{z + 2}) = \frac{π}{4}

, then the minimum value of

∣ z - 92 - 2 i ∣^{2}

is equal to ____.

Answer: 98

The condition

ar g (\frac{z - 2}{z + 2}) = \frac{π}{4}

puts

z

on an arc through

(2, 0)

and

(- 2, 0)

on which that chord subtends

\frac{π}{4}

. The central angle is

\frac{π}{2}

, so

4 = 2 R sin \frac{π}{4}

gives

R = 22

, and the centre lies on the

y

-axis at

(0, 2)

. The point

92 + 2 i = (92, 2)

is at distance

92

from the centre, so its least distance to the arc is

92 - 22 = 72

. Thus the minimum of

∣ z - 92 - 2 i ∣^{2}

(72)^{2} = 98

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer