Question

Let $\omega_1$ and $\omega_2$ are complex numbers such that $\left|\frac{\omega_1-1}{\omega_1-4}\right|=2$ and $\left|\frac{\omega_2-4}{\omega_2-1}\right|=2$. If $|\omega_1-\omega_2|_{max}=\alpha|\omega_1-\omega_2|_{min}$ (where $|\omega_1-\omega_2|_{max}$ and $|\omega_1-\omega_2|_{min}$ denote maximum and minimum value of $|\omega_1-\omega_2|$ respectively). Then $\alpha$ equals to

• A. 2
• B. 10
• C. 9
• D. 11

Answer 1

Answer: (C)

9

Answer 2

The locus of points $z$ satisfying $\left|\frac{z-a}{z-b}\right|=k$ where $k \neq 1$ is a circle (Circle of Apollonius).

1. Locus of $\omega_1$: Given $\left|\frac{\omega_1-1}{\omega_1-4}\right|=2$. This implies $|\omega_1-1| = 2|\omega_1-4|$. Squaring both sides: $|\omega_1-1|^2 = 4|\omega_1-4|^2$. Let $\omega_1 = x+iy$. $(x-1)^2 + y^2 = 4((x-4)^2 + y^2)$ $x^2 - 2x + 1 + y^2 = 4(x^2 - 8x + 16 + y^2)$ $3x^2 - 30x + 63 + 3y^2 = 0$ $x^2 - 10x + 21 + y^2 = 0$ Completing the square for $x$: $(x-5)^2 + y^2 = 4$. This is a circle $C_1$ with center $c_1 = 5$ and radius $r_1 = 2$.

2. Locus of $\omega_2$: Given $\left|\frac{\omega_2-4}{\omega_2-1}\right|=2$. This implies $|\omega_2-4| = 2|\omega_2-1|$. Squaring both sides: $|\omega_2-4|^2 = 4|\omega_2-1|^2$. Let $\omega_2 = x+iy$. $(x-4)^2 + y^2 = 4((x-1)^2 + y^2)$ $x^2 - 8x + 16 + y^2 = 4(x^2 - 2x + 1 + y^2)$ $3x^2 - 12 + 3y^2 = 0$ $x^2 + y^2 = 4$. This is a circle $C_2$ with center $c_2 = 0$ and radius $r_2 = 2$.

3. Distance between points on the circles: The distance between the centers of the circles is $d = |c_1 - c_2| = |5 - 0| = 5$. The minimum distance between points on $C_1$ and $C_2$ is $|\omega_1 - \omega_2|_{min} = d - r_1 - r_2 = 5 - 2 - 2 = 1$. The maximum distance between points on $C_1$ and $C_2$ is $|\omega_1 - \omega_2|_{max} = d + r_1 + r_2 = 5 + 2 + 2 = 9$.

4. Finding $\alpha$: We are given $|\omega_1-\omega_2|_{max}=\alpha|\omega_1-\omega_2|_{min}$. Substituting the values: $9 = \alpha \times 1$, so $\alpha = 9$.