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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Complex Number

Let | z -i 2 z +i |=(1)/(3), z ∈ C , be the equation of a circle...

Let ∣zˉ−i2zˉ+i∣=13,z∈C\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}, z \in C, be the equation of a circle with center at CC. If the area of the triangle, whose vertices are at the points (0,0),Cand(α,0)(0,0), \mathrm{C} and (\alpha, 0) is 11 square units, then α2\alpha^2 equals:

Options

  1. A5050
  2. B100100
  3. C8125\frac{81}{25}
  4. D12125\frac{121}{25}

Correct answer

B. 100100

Step-by-step solution

Let z=x+iy⇒zˉ=x−iyz=x+i y \Rightarrow \bar{z}=x-i y3∣zˉ−i∣=1∣2zˉ+i∣=3∣(x−(y+1)i∣=∣2x+i(1−2y)∣=3x2+(y+1)2=(2x)2+(1−2y)2=9(x2+y2+2y+1)=4x2+4y2−4y+1⇒5x2+5y2+22y+8=0⇒ Centre ≡(0,−115)\begin{aligned} & 3|\bar{z}-i|=1|^{2} \bar{z}+i| \\ & =3 \mid(x-(y+1) i|=|^{2} x+i(1-2 y)| \\ & =3 \sqrt{x^2+(y+1)^2}=\sqrt{(2 x)^2+(1-2 y)^2} \\ & =9\left(x^2+y^2+2 y+1\right)=4 x^2+4 y^2-4 y+1 \\ & \Rightarrow 5 x^2+5 y^2+22 y+8=0 \\ & \Rightarrow \text{ Centre } \equiv\left(0,-\frac{11}{5}\right) \end{aligned}Area of $\Delta =\frac{1}{2}|\alpha|\left|\frac{-11}{5}\right|=11\begin{aligned} & \Rightarrow \quad|\alpha|=10 \\ & \Rightarrow \quad \alpha^2=100\end{aligned}$$$
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