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

The equation of the circles are
`|z-z_(0)|=r`……………(i) and `|z-z_(0)|=2r` …………….(ii)
`therefore |alpha-z_(0)|=r^(2)` and `|1/alpha-z_(0)|^(2)=4r^(2)`
`rArr |alpha-z_(0)|^(2)=r^(2)` and `|alpha/|alpha|^(2)-z_(0)|^(2)=4r^(2)`
`rArr (alpha-z_(0))|^(2)=4r^(2)`
`rArr (alpha-z_(0))(baralpha-barz_(0))=r^(2)` and `(alpha/|alpha|^(2)-z_(0))(baralpha/|(alpha)|^(2)-barz_(0))=4r^(2)`
`rArr alphabaralpha=alphabarz_(0)-baralphaz_(0)+z_(0)barz_(0)=r^(2)`
and `, (alphabaralpha)/(|alpha|^(4))-(alphabarz_(0))/(|alpha|^(2))-(baralphaz_(0))/(|alpha|^(2))|+z_(0)barz_(0)=4r^(2)`
`rArr |alpha|^(2)-alphabarz_(0)-baralphaz_(0)+|z_(0)|^(2)=4r^(2)`
`rArr |alpha|^(2)-(alphabarz_(0)+baralphaz_(0))+|z_(0)|^(2)`
and `1/|a|^(2)-1/|alpha|^(2)(alphabarz_(0)+baralphaz_(0))+|z_(0)|^(2)=4r^(2)`
`rArr |alpha|^(2)-(alphabarz_(0)-baraz_(0))+|z_(0)|^(2)=r^(2)`
and,` 1-(abarz_(0)+baralphaz_(0))+|alpha|^(2)|z_(0)|^(2)=4|alpha|^(2)r^(2)`
Subtracting first equation from second equation, we get
`(1-|alpha|^(2))+(|alpha|^(2)-1)|z_(0)|^(2)=(4|alpha|^(2)-1)r^(2)`
`rArr 1-|alpha|^(2)-1=8|alpha|^(2) -2 rArr 7|alpha|^(2) rArr |alpha|=1/sqrt(7)`