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 \(|z - z_1| = k |z - z_2|\) for \(k \ne 1\) represent a circle. Square and simplify using coordinate geometry.

Answer 1

The Correct Option is A

Let \(z = x + iy\), Then we have: \[ |z - z_1| = |x - 1 + i(y - 2)| = \sqrt{(x - 1)^2 + (y - 2)^2} \] \[ |z - z_2| = |x + i(y - 3)| = \sqrt{x^2 + (y - 3)^2} \] Given: \[ \sqrt{(x - 1)^2 + (y - 2)^2} = 2\sqrt{x^2 + (y - 3)^2} \] Squaring both sides: \[ (x - 1)^2 + (y - 2)^2 = 4(x^2 + (y - 3)^2) \] LHS: \[ x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 - 2x - 4y + 5 \] RHS: \[ 4(x^2 + y^2 - 6y + 9) = 4x^2 + 4y^2 - 24y + 36 \] Now subtract RHS from LHS: \[ x^2 + y^2 - 2x - 4y + 5 - 4x^2 - 4y^2 + 24y - 36 = 0 \Rightarrow -3x^2 - 3y^2 - 2x + 20y - 31 = 0 \] Divide by \(-3\): \[ x^2 + y^2 + \frac{2x}{3} - \frac{20y}{3} + \frac{31}{3} = 0 \] Complete the square: For \(x^2 + \frac{2x}{3}\): \[ = \left(x + \frac{1}{3}\right)^2 - \frac{1}{9} \] For \(y^2 - \frac{20y}{3}\): \[ = \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} \] Substitute: \[ \left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} = -\frac{31}{3} \Rightarrow \left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = \frac{8}{9} \] This is a circle with: - Center: \(\left(-\frac{1}{3},\ \frac{10}{3}\right)\) ✅ - Radius: \(\sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\) ✅