Question

The locus of the point \(z=x+iy\) satisfying \[ \left|\frac{z-(2+i)}{z+(2-i)}\right|=2 \] is:

• A. A circle
• B. A parabola
• C. An ellipse
• D. A straight line

Hint:

A locus of the form \[ \left|\frac{z-z_1}{z-z_2}\right|=k \] represents an Apollonius circle whenever \(k\neq1\). For \(k=1\), the locus becomes the perpendicular bisector of the segment joining the fixed points.

Answer 1

The Correct Option is A

Concept: Expressions of the form \[ \left|\frac{z-z_1}{z-z_2}\right|=k \] represent the ratio of distances from two fixed points in the Argand plane. Such loci are known as Apollonius circles whenever \(k\neq1\).

Step 1: Rewrite the modulus equation Given \[ \left|\frac{z-(2+i)}{z+(2-i)}\right|=2. \] Taking modulus separately, \[ |z-(2+i)| = 2|z+(2-i)|. \]

Step 2: Interpret geometrically The point \(z=x+iy\) has distances \[ |z-(2+i)| \] from the fixed point \[ (2,1) \] and \[ |z+(2-i)| = |z-(-2,1)| \] from the fixed point \[ (-2,1). \] Thus the ratio of distances from two fixed points is constant: \[ \frac{\text{Distance from }(2,1)} {\text{Distance from }(-2,1)} =2. \]

Step 3: Use the standard result The locus of a point whose distances from two fixed points have a constant ratio different from \(1\) is an Apollonius circle. Hence the locus is a circle. \[ \boxed{\text{Circle}} \]

Step 4: Verification algebraically Let \[ z=x+iy. \] Then \[ \sqrt{(x-2)^2+(y-1)^2} = 2\sqrt{(x+2)^2+(y-1)^2}. \] Squaring and simplifying yields a second-degree equation of a circle. Hence the conclusion is confirmed.