Question

Let a complex number \(z\) satisfy the equation: \[ |z - z_1| = 2 |z - z_2|,\quad \text{where } z_1 = 1 + 2i,\ z_2 = 3i \] Then choose the correct options:

• A. Center of circle is \(\left(-\frac{1}{3},\ \frac{10}{3}\right)\)
• B. Center of circle is \(\left(\frac{1}{3},\ -\frac{10}{3}\right)\)
• C. Radius of circle is \(\frac{2\sqrt{2}}{3}\)
• D. Radius of circle is \(\frac{3\sqrt{2}}{4}\)

Hint:

Equations of the form \(|z - z_1| = k|z - z_2|\) represent a circle (if \(k \ne 1\)). Use coordinate geometry to simplify.

Answer 1

The Correct Option is A

Let \(z = x + iy\) Then: \[ |z - z_1| = \sqrt{(x - 1)^2 + (y - 2)^2}, \quad |z - z_2| = \sqrt{x^2 + (y - 3)^2} \] Given: \[ \sqrt{(x - 1)^2 + (y - 2)^2} = 2 \sqrt{x^2 + (y - 3)^2} \] Square both sides: \[ (x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2) \] Expand both sides: LHS: \[ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 - 2x - 4y + 5 \] RHS: \[ 4x^2 + 4(y^2 - 6y + 9) = 4x^2 + 4y^2 - 24y + 36 \] Now equate: \[ x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36 \] Bring all terms to one side: \[ x^2 + y^2 - 2x - 4y + 5 - 4x^2 - 4y^2 + 24y - 36 = 0 \Rightarrow -3x^2 -3y^2 - 2x + 20y - 31 = 0 \] Divide entire equation by \(-3\): \[ x^2 + y^2 + \frac{2x}{3} - \frac{20y}{3} + \frac{31}{3} = 0 \] Complete the square: \[ x^2 + \frac{2x}{3} + y^2 - \frac{20y}{3} = -\frac{31}{3} \] Complete: - For \(x\): \(x^2 + \frac{2x}{3} = \left(x + \frac{1}{3}\right)^2 - \frac{1}{9}\) - For \(y\): \(y^2 - \frac{20y}{3} = \left(y - \frac{10}{3}\right)^2 - \frac{100}{9}\) Substitute back: \[ \left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} = -\frac{31}{3} \] Combine: \[ \left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = -\frac{31}{3} + \frac{101}{9} = \frac{-93 + 101}{9} = \frac{8}{9} \] Thus: - Center: \(\left(-\frac{1}{3},\ \frac{10}{3}\right)\) ✅ - Radius: \(\sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\) ✅