If the locus of \(z \in \mathrm{C}\), such that \[ \operatorname{Re}\left(\frac{z-1}{2 z+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-\mathrm{i}}\right)=2, \] is a circle of radius \(r\) and center \((a, b)\) then \(\frac{15 a b}{r^{2}}\) is equal to :

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Complex Numbers

If the locus of

z \in C

, such that

Re (\frac{z - 1}{2 z + i}) + Re (\frac{z ˉ - 1}{2 z ˉ - i}) = 2,

is a circle of radius

r

and center

(a, b)

then

\frac{15 ab}{r ^{2}}

is equal to :

Answer: (C)

Because

\frac{z ˉ - 1}{2 z ˉ - i}

is the conjugate of

\frac{z - 1}{2 z + i}

, the two real parts are equal, so the condition reduces to

Re (\frac{z - 1}{2 z + i}) = 1

. Put

z = x + i y

; the real part is

\frac{2 x ( x - 1 ) + y ( 2 y + 1 )}{4 x ^{2} + ( 2 y + 1 ) ^{2}} = 1

. Cross-multiplying and simplifying gives

2 x^{2} + 2 y^{2} + 2 x + 3 y + 1 = 0

, i.e.

x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0

. The centre is

(- \frac{1}{2}, - \frac{3}{4})

and

r^{2} = \frac{1}{4} + \frac{9}{16} - \frac{1}{2} = \frac{5}{16}

. Therefore

\frac{15 ab}{r ^{2}} = 15 \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{16}{5} = 18

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option