JEE MAIN - Mathematics (2023 - 12th April Morning Shift - No. 6)
Let $$\mathrm{C}$$ be the circle in the complex plane with centre $$\mathrm{z}_{0}=\frac{1}{2}(1+3 i)$$ and radius $$r=1$$. Let $$\mathrm{z}_{1}=1+\mathrm{i}$$ 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 :
$$\frac{3}{2}$$
$$\frac{5}{2}$$
$$\frac{13}{2}$$
$$\frac{7}{2}$$
Explanation
Given, $$
z_0=\frac{1+3 i}{2}, z_1=(1+i)
$$
$$ \left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}} $$
$$ \begin{aligned} & \left|z_1-z_0\right|\left|z_2-z_0\right|=1 \\\\ & \Rightarrow \frac{1}{\sqrt{2}}\left|z_2-z_0\right|=1 \\\\ & \Rightarrow\left|z_2-z_0\right|=\sqrt{2} \end{aligned} $$
$$ \begin{aligned} & \frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}( \pm 1)= \pm 2 \\\\ & z_2=z_0 \pm 2\left(z_1-z_0\right) \end{aligned} $$
$$ z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2} $$
OR
$$ z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2} $$
$$ \left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}} $$
$$ \begin{aligned} & \left|z_1-z_0\right|\left|z_2-z_0\right|=1 \\\\ & \Rightarrow \frac{1}{\sqrt{2}}\left|z_2-z_0\right|=1 \\\\ & \Rightarrow\left|z_2-z_0\right|=\sqrt{2} \end{aligned} $$
$$ \begin{aligned} & \frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}( \pm 1)= \pm 2 \\\\ & z_2=z_0 \pm 2\left(z_1-z_0\right) \end{aligned} $$
$$ z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2} $$
OR
$$ z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2} $$
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