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If $|z_1+2+2i|\leq \frac {2}{3},|z_2-1-2i|\leq \frac {3}{2} $ then minimum value of $|z_2-z_1|$ is? .$$ \text{Attempt} $$ Now I know that the two equations represent all set of circles having radii less than or equal to $2/3,3/2$ respectively centred at $(-2,-2) (1,2) $ So I can find the equation of line joining their centres and then get the points which are nearer to each other. But that seems to be very lengthy procedure. Also using $|z_1-z_2-z_3|\geq| |z_1|-|z_2|-|z_3||$ I get wrong answer. Is there any better approach for this problem . Thanks!

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The distance between their centers of the circles is $$ \sqrt{(-2-1)^2+(-2-(-2))^2}=5. $$ The radius of the first circle is $2/3$. The radius of the second is $3/2$. Hence (since $5>2/3+3/2$), the minimum distance between the circles is

$$5-(2/3+3/2)=\frac{17}{6}.$$

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  • $\begingroup$ The options given are $5,\frac {17}{6},\frac {49}{6},\frac {13}{3} $ $\endgroup$ – Archis Welankar Apr 16 '17 at 17:03
  • $\begingroup$ Sorry, I messed up a minus sign. I think it should be corrected now. The idea is exactly the same, though. $\endgroup$ – mickep Apr 16 '17 at 17:09

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