Question

Let \( z \) be a complex number such that \( |z-6|=5 \) and \( |z+2-6i|=5 \). Then the value of \( z^3+3z^2-15z+14 \) is equal to:

• A. \(37\)
• B. \(42\)
• C. \(50\)
• D. \(61\)

Hint:

When two circles with equal radii touch externally, the common point lies exactly at the midpoint of their centers.

Answer 1

The Correct Option is A

Concept: The given conditions represent two circles in the Argand plane. If two circles with equal radii touch externally, the point of contact is the midpoint of their centers.
Step 1: Identify the centers of the circles \[ |z-6|=5 \Rightarrow \text{Center } C_1=(6,0) \] \[ |z+2-6i|=5 \Rightarrow \text{Center } C_2=(-2,6) \] Distance between centers: \[ \sqrt{(6+2)^2+(0-6)^2}=\sqrt{64+36}=10 \] Since distance \(=\) sum of radii, the circles touch externally.
Step 2: Find the point of contact \[ z=\left(\frac{6+(-2)}{2},\frac{0+6}{2}\right)=(2,3) \Rightarrow z=2+3i \]
Step 3: Evaluate the given expression \[ z^2=(2+3i)^2=-5+12i \] \[ z^3=(2+3i)(-5+12i)=-46+9i \] \[ z^3+3z^2-15z+14 =(-46+9i)+3(-5+12i)-15(2+3i)+14 \] \[ =(-46-15-30+14)+(9+36-45)i =37 \]