InterviewSolution
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InterviewSolution
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Let complex numbers `alpha and 1/alpha` lies on circle `(x-x_0)^2(y-y_0)^2=r^2 and (x-x_0)^2+(y-y_0)^2=4r^2` respectively. If `z_0=x_0+iy_0` satisfies the equation `2|z_0|^2=r^2+2` then `|alpha|` is equal to (a) `1/sqrt2` (b) `1/2` (c) `1/sqrt7` (d) `1/3`A. `1//sqrt(2)`B. `1//2`C. `1//sqrt(7)`D. `1//3` |
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Answer» Correct Answer - C Given circles are `(x - x_(0))^(2) + (y-y_(0))^(2) = r^(2)` and `(x-x_(0))^(2) + (y - y_(0))^(2) = 4r^(2)" "(1)` or `|z-z_(0)| = r` and `|z-z_(0)| = 2 r` where `z_(0) - x_(0) + iy_(0)` Now `alpha` and `(1)/(baralpha)` lies on circle (1) and (2), respectively. Then ` |alpha -z_(0)| = r and |(1)/(baralpha) - z_(0)| = 2r` `rArr |alpha - z_(0)| = r and |1-baralphaz_(0)| = 2r |baralpha|` `rArr |alpha -z_(0)| = r and |1-baraz_(0)| = 2r|baralpha|` `rArr |alpha - z_(0)|^(2) =r^(2) and |1-baralphaz_(0)| = 4r^(2) |alpha|^(2)` Subtracting, we get `|1-baralphaz_(0)|^(2) - |alpha -z_(0)|^(2) = 4r^(2) |alpha| - r^(2)` `rArr 1+ |alphaz_(0)|^(2) -baralphaz_(0) - alphabarz_(0) -(|alpha|^(2) + |z_(0)|^(2) -baralphaz_(0) -alphabarz_(0))` `4r^(2) |alpha|^(2) -r^(2)` `rArr 1+|alpha|^(2) |z_(0)|^(2) - |alpha|^(2) -|z_(0)|^(2) = 4r^(2)|alpha|^(2)-r^(2)` Given ` 2|z_(0)|^(2) = r^(2) + 2` `2|z_(0)|^(2) = r^(2)+ 2` `rArr (1-|alpha|^(2))(1-(r^(2) +2)/(2)) = 4r^(2)|alpha|^(2) -r^(2)` `rArr (1-|alpha|^(2))((-r^(2))/(2))= 4r^(2)|alpha|^(2) -r^(2)` `rArr |alpha|^(2) -1 = 8|alpha|^(2)-2` `rArr |alpha|^(2) =(1)/(7) rArr |alpha| = (1)/(sqrt(7))` |
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