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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

z = x + iy

, then the equation

\left|\dfrac{2z - i}{z + 1}\right| = m

does not represent a circle when

m = \dfrac{1}{2}

m = 1

m = 2

correct

m = 3

Solution

Step 1: Clear the modulus. The condition

\left|\dfrac{2z - i}{z + 1}\right| = m

is equivalent to

|2z - i| = m|z + 1|

. Square both sides (both are non-negative):

|2z - i|^2 = m^2 |z + 1|^2

Step 2: Expand each modulus squared using

|w|^2 = w\bar{w}

and

z = x + iy

|z|^2 = x^2 + y^2

. Left side:

|2z - i|^2 = (2x)^2 + (2y - 1)^2 = 4x^2 + 4y^2 - 4y + 1 = 4(x^2 + y^2) - 4y + 1

Right side:

m^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(x^2 + y^2) - 4y + 1 - m^2(x^2 + y^2) - 2m^2 x - m^2 = 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(x^2 + y^2) + Dx + Ey + F = 0

represents a circle only when

A \neq 0

. The coefficient of

x^2 + y^2

here is

A = 4 - m^2

. It represents a circle for every

m

except when:

4 - m^2 = 0 \implies m^2 = 4 \implies m = 2 \quad (m > 0)

Step 5: When

m = 2

, the quadratic terms vanish and the equation becomes linear (

-8x - 4y - 3 = 0

), a straight line, not a circle. Correct answer: (3)

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