Let C be the circle in the complex plane with centre z 0 = 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 | z 1 - z 0 | | z 2 - z 0 | = 1 . If z 0 , z 1 , and z 2 are collinear

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Solution

Q.
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 $| z_{1} - z_{0} |$ $| z_{2} - z_{0} |$ $= 1 .$ If $z_{0}, z_{1}$ , and $z_{2}$ are collinear, then the smaller value of $| z_{2} |^{2}$ is equal to [2023]

1 $\frac{5}{2}$

2 $\frac{7}{2}$

3 $\frac{3}{2}$

4 $\frac{13}{2}$

Ans.
(1)

$z_{0} = (\frac{1}{2} + \frac{3 i}{2})$

$z_{1} = 1 + i$

$z_{1} - z_{0} = (1 + i) - (\frac{1}{2} + \frac{3 i}{2}) = \frac{1}{2} - \frac{i}{2} = \frac{1 - i}{2}$

$| z_{1} - z_{0} |$ $= \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$

$As | z_{1} - z_{0} |$ $| z_{2} - z_{0} |$ $= 1$

$\Rightarrow \frac{1}{\sqrt{2}}$ $| z_{2} - z_{0} |$ $= 1 \Rightarrow$ $| z_{2} - z_{0} |$ $= \sqrt{2}$

Slope of line forming $z_{0}, z_{2} =$ $\frac{\frac{3}{2} - 1}{\frac{1}{2} - 1} = \frac{\frac{1}{2}}{\frac{- 1}{2}} = - 1$

$\Rightarrow \tan θ = - 1 = \tan 135 ° \Rightarrow θ = 135 °$

$| z_{2} - z_{0} |$ $= \sqrt{2}$

$z_{2} (\frac{1}{2} + \sqrt{2} \cos 135 °, \frac{3}{2} + \sqrt{2} \sin 135 °)$

or

$z_{2} (\frac{1}{2} - \sqrt{2} \cos 135 °, \frac{3}{2} - \sqrt{2} \sin 135 °)$

$z_{2} (\frac{- 1}{2}, \frac{5}{2})$ or $z_{2} (\frac{3}{2}, \frac{1}{2})$

$| z_{2} |^{2} = \frac{1}{4} + \frac{25}{4} = \frac{26}{4}$ or $| z_{2} |^{2} = \frac{9}{4} + \frac{1}{4} = \frac{5}{2};$ $\Rightarrow | z_{2} |_{\min}^{2} = \frac{5}{2}$

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