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

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$\frac{5}{2}$

$\frac{7}{2}$

$\frac{13}{2}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$S = {(x - \frac{1}{2})}^{2} + {(y - \frac{3}{2})}^{2} = 1$
$Z_{1}$ = (1,1) lies inside S and $Z_{2}$ lies outside S
$| Z_{1} - Z_{0} | = \frac{1}{\sqrt{2}}$
Given, $| Z_{1} - Z_{0} | | Z_{2} - Z_{0} | = 1$
$∴ | Z_{2} - Z_{0} | = \sqrt{2} \Rightarrow {(x - \frac{1}{2})}^{2} + {(y - \frac{3}{2})}^{2} = 2 . . . . (1)$
Equation of the line joining $Z_{0}, Z_{1}, Z_{2}$ is x + y = 2----------(2)
On solving (1) & (2), we have
$Z_{2} = (\frac{3}{2}, \frac{1}{2})$ or $(\frac{- 1}{2}, \frac{5}{2})$
$| Z_{2} | = \sqrt{\frac{5}{2}}$ or $\sqrt{\frac{13}{2}}$
$∴$ Smaller value of ${| Z_{2} |}^{2} = \frac{5}{2}$