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When |2z−i|/|z+1|=m Is Not a Circle | Complex Locus JEE

JEE Maths question with a full step-by-step solution.

Question
If z=x+iyz = x + iy, then the equation ∣2z−iz+1∣=m\left|\dfrac{2z - i}{z + 1}\right| = m does not represent a circle when
Am=12m = \dfrac{1}{2}
Bm=1m = 1
Cm=2m = 2correct
Dm=3m = 3
Solution
Step 1: Clear the modulus. The condition ∣2z−iz+1∣=m\left|\dfrac{2z - i}{z + 1}\right| = m is equivalent to ∣2z−i∣=m∣z+1∣|2z - i| = m|z + 1|. Square both sides (both are non-negative):
∣2z−i∣2=m2∣z+1∣2|2z - i|^2 = m^2 |z + 1|^2
 Step 2: Expand each modulus squared using ∣w∣2=wwˉ|w|^2 = w\bar{w} and z=x+iyz = x + iy, ∣z∣2=x2+y2|z|^2 = x^2 + y^2. Left side:
∣2z−i∣2=(2x)2+(2y−1)2=4x2+4y2−4y+1=4(x2+y2)−4y+1|2z - i|^2 = (2x)^2 + (2y - 1)^2 = 4x^2 + 4y^2 - 4y + 1 = 4(x^2 + y^2) - 4y + 1
 Right side:
m2∣z+1∣2=m2[(x+1)2+y2]=m2(x2+y2)+2m2x+m2m^2|z + 1|^2 = m^2\big[(x + 1)^2 + y^2\big] = m^2(x^2 + y^2) + 2m^2 x + m^2
 Step 3: Bring everything to one side:
4(x2+y2)−4y+1−m2(x2+y2)−2m2x−m2=04(x^2 + y^2) - 4y + 1 - m^2(x^2 + y^2) - 2m^2 x - m^2 = 0
(4−m2)(x2+y2)−2m2x−4y+(1−m2)=0(4 - m^2)(x^2 + y^2) - 2m^2 x - 4y + (1 - m^2) = 0
 Step 4: A second-degree equation of the form A(x2+y2)+Dx+Ey+F=0A(x^2 + y^2) + Dx + Ey + F = 0 represents a circle only when A≠0A \neq 0. The coefficient of x2+y2x^2 + y^2 here is A=4−m2A = 4 - m^2. It represents a circle for every mm except when:
4−m2=0  ⟹  m2=4  ⟹  m=2(m>0)4 - m^2 = 0 \implies m^2 = 4 \implies m = 2 \quad (m > 0)
 Step 5: When m=2m = 2, the quadratic terms vanish and the equation becomes linear (−8x−4y−3=0-8x - 4y - 3 = 0), a straight line, not a circle. Correct answer: (3)
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