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:
- A \(50\)
- B \(100\)
- C \(\frac{81}{25}\)
- D \(\frac{121}{25}\)
Answer & Solution
Correct Answer
(B) \(100\)
Step-by-step Solution
Detailed explanation
Let \(z=x+i y \Rightarrow \bar{z}=x-i y\)…
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