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JEE Mains · Maths · STD 11 - 4.1 complex nubers

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

  1. A \(50\)
  2. B \(100\)
  3. C \(\frac{81}{25}\)
  4. D \(\frac{121}{25}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(100\)

Step-by-step Solution

Detailed explanation

Let \(z=x+i y \Rightarrow \bar{z}=x-i y\)…