Let C be the circle in the complex plane with centre $\mathrm{z}_{0}=\frac{1}{2}(1+3 \mathrm{i})$ and radius $\mathrm{r}=1$. Let $\mathrm{z}_{1}=1+\mathrm{i}$ and the complex number $\mathrm{z}_{2}$ be outside the circle C such that $\left|\mathrm{z}_{1}-\mathrm{z}_{0}\right|\left|\mathrm{z}_{2}-\mathrm{z}_{0}\right|=1$. If $\mathrm{z}_{0}, \mathrm{z}_{1}$ and $\mathrm{z}_{2}$ are collinear, then the smaller value of $\left|z_{2}\right|^{2}$ is equal to

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Complex Numbers

Let C be the circle in the complex plane with centre

z_{0} = \frac{1}{2} (1 + 3 i)

and radius

r = 1

. Let

z_{1} = 1 + i

and the complex number

z_{2}

be outside the circle C such that

∣ z_{1} - z_{0} ∣ ∣ z_{2} - z_{0} ∣ = 1

. If

z_{0}, z_{1}

and

z_{2}

are collinear, then the smaller value of

∣ z_{2} ∣^{2}

is equal to

Answer: (B)

We have

∣ z_{1} - z_{0} ∣ = \frac{1}{2} - \frac{1}{2} i = \frac{1}{2}

. From the condition

∣ z_{1} - z_{0} ∣∣ z_{2} - z_{0} ∣ = 1

, we get

∣ z_{2} - z_{0} ∣ = 2

. Since

z_{0}, z_{1}, z_{2}

are collinear, the line through

z_{0} = (\frac{1}{2}, \frac{3}{2})

and

z_{1} = (1, 1)

has slope

- 1

, making an angle of

13 5^{\circ}

with the positive x-axis. The two points at distance

2

from

z_{0}

along this line are

z_{2} = (- \frac{1}{2}, \frac{5}{2})

and

z_{2} = (\frac{3}{2}, \frac{1}{2})

. Their moduli squared are

\frac{1}{4} + \frac{25}{4} = \frac{26}{4}

and

\frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}

. The smaller value is

\frac{5}{2}

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option