Jee Mains
Let C be the circle in the complex plane with centre $\mathrm{z}_{0}=\frac{1}{2}(1+3 \mathrm{i})$ and radius r = 1. Let $\mathrm{z}_{1}=1+\mathrm{i}$ and the complex number $\mathrm{z}_{2}$ be outside the circle C such that $\left|\mathrm{z}_{1}-\mathrm{z}_{0}\right|\left|\mathrm{z}_{2}-\mathrm{z}_{0}\right|=1$.
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Solution
According to the question,
\[ \begin{aligned} & \left|z_{1}-z_{0}\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}} \Rightarrow\left|z_{1}-z_{0}\right|=\sqrt{2} ; \text { centre }\left(\frac{1}{2}, \frac{3}{2}\right) \\ & z_{0}\left(\frac{1}{2}, \frac{3}{2}\right) \text { and } z_{1}(1,1) \Rightarrow \tan \theta=-1 \Rightarrow \theta=135^{\circ} \\ & z_{2}\left(\frac{1}{2}+\sqrt{2} \cos 135^{\circ}, \frac{3}{2}+\sqrt{2} \sin 135^{\circ}\right) \text { or }\left(\frac{1}{2}-\sqrt{2} \cos 135^{\circ}, \frac{3}{2}-\sqrt{2} \sin 135^{\circ}\right) \\ & \Rightarrow z_{2}\left(-\frac{1}{2}, \frac{5}{2}\right) \text { or } z_{2}\left(\frac{3}{2}, \frac{1}{2}\right) \Rightarrow\left|z_{2}\right|^{2}=\frac{26}{4}, \frac{5}{2} \Rightarrow\left|z_{2}\right|_{\min }^{2}=\frac{5}{2} \end{aligned} \]
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