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

Options

A. `1//sqrt(2)`
B. `1//2`
C. `1//sqrt(7)`
D. `1//3`

Answer

C

Explanation

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))`