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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(C\) be the circle in the complex plane with centre \(z_0=\frac{1}{2}(1+3 i)\) and radius \(r=1\). Let \(z_1=1+\) \(i\) and the complex number \(z_2\) be outside the circle \(C\) such that \(\left|z_1-z_0\right|\left|z_2-z_0\right|=1\). If \(z_0, z_1\) and \(z_2\) are collinear, then the smaller value of \(\left|z_2\right|^2\) is equal to \(.............\). 

  1. A \(\frac{13}{2}\)
  2. B \(\frac{5}{2}\)
  3. C \(\frac{3}{2}\)
  4. D \(\frac{7}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{5}{2}\)

Step-by-step Solution

Detailed explanation

\(\left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}}\) \(\Rightarrow\left|z_2-z_{\circ}\right|=\sqrt{2} ; \text { centre }\left(\frac{1}{2}, \frac{3}{2}\right)\) \(z_{0}\left(\frac{1}{2}, \frac{3}{2}\right) \text { and } z_1(1,1)\)…
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