Let $S=\{z \in \mathbb{C}:|z-3| \leq 1$ and $z(4+3 i)+\bar{z}(4-3 i) \leq 24\}$. If $\alpha+i \beta$ is the point in S which is closest to $4 i$, then $25(\alpha+\beta)$ is equal to ______.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Complex Numbers

Let

S = {z \in C : ∣ z - 3∣ \leq 1

and

z (4 + 3 i) + \overset{z}{ˉ} (4 - 3 i) \leq 24}

.
If

α + i β

is the point in S which is closest to

4 i

, then

25 (α + β)

is equal to ______.

Answer: 80

The condition

∣ z - 3∣ \leq 1

defines a disk of radius 1 centered at

(3, 0)

. Writing

z = x + i y

z (4 + 3 i) + \overset{z}{ˉ} (4 - 3 i) = 8 x - 6 y \leq 24

, or

4 x - 3 y \leq 12

. The point in

S

closest to

(0, 4)

lies on the line joining

(0, 4)

and

(3, 0)

\frac{x}{3} + \frac{y}{4} = 1

, or

4 x + 3 y = 12

. Substituting into the circle equation

(x - 3)^{2} + y^{2} = 1

(\frac{12 - 3 y}{4} - 3)^{2} + y^{2} = 1 \Rightarrow \frac{9 y ^{2}}{16} + y^{2} = 1 \Rightarrow y^{2} = \frac{16}{25}

. Taking

y = \frac{4}{5}

(closer to 4), we get

x = \frac{12}{5}

. Thus

25 (α + β) = 25 (\frac{12}{5} + \frac{4}{5}) = 25 \cdot \frac{16}{5} = 80

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer