Question 7

MCQMEDIUM

Let $\mathbb{R}$ denote the set of all real numbers. Let $z_1 = 1 + 2i$ and $z_2 = 3i$ be two complex numbers, where $i = \sqrt{-1}$ . Let $S = \{ (x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2| \}.$ Then which of the following statements is (are) TRUE?

(A)

$S$ is a circle with centre $(-\frac{1}{3}, \frac{10}{3})$

(B)

$S$ is a circle with centre $(\frac{1}{3}, \frac{8}{3})$

(C)

$S$ is a circle with radius $\frac{\sqrt{2}}{3}$

(D)

$S$ is a circle with radius $\frac{2\sqrt{2}}{3}$

Detailed Solution

Let $z = x + iy$ . The condition $|z - z_1| = 2|z - z_2|$ implies $|z - (1 + 2i)|^2 = 4|z - 3i|^2$ .

This expands to $(x-1)^2 + (y-2)^2 = 4[x^2 + (y-3)^2]$ .

Expanding and simplifying: $x^2 - 2x + 1 + y^2 - 4y + 4 = 4x^2 + 4y^2 - 24y + 36 \Rightarrow 3x^2 + 3y^2 + 2x - 20y + 31 = 0$ .

Dividing by 3 gives $x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0$ .

The centre is $(-\frac{1}{3}, \frac{10}{3})$ and the radius $r = \sqrt{(\frac{1}{3})^2 + (\frac{10}{3})^2 - \frac{31}{3}} = \sqrt{\frac{1 + 100 - 93}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$ .

Thus (A) and (D) are correct.

Question 7

Detailed Solution

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