Complex Numbers
Algebra of Complex Numbers
Grade Class 11
Question:
<p>Let \(z\) satisfy \(|z+16|=4|z+1|\). Which are TRUE?</p>
\(|z|=4\) when \(z\) is real
\(z\) lies on circle of radius 4
\(|z+4|^2 = |z|^2+4\)
Circle passes through origin
Step-by-Step Solution
Key Concept: |z+16|=4|z+1| is an Apollonius circle. Squaring: (x+16)^2+y^2=16((x+1)^2+y^2) \Rightarrow x^2+y^2+32x+256 = 16x^2+32x+16+16y^2. This simplifies to a circle equation.
<p>Expand: \(x^2+32x+256+y^2 = 16x^2+32x+16+16y^2\). \( \Rightarrow 15x^2+15y^2=240 \Rightarrow x^2+y^2=16\). Circle of radius 4 centred at origin, which passes through \((\pm 4,0)\) but not through \((0,0)\) — D is TRUE since \(x^2+y^2=16\neq 0\). Check again: origin is NOT on the circle. So only B is correct.</p>
Correct Answer: BD