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

Formula used
`|z|^2=z. bar z`
and `|z_1=z_2|^2=(z_1-z_2)(barz_1-z_2)`
`=|z_1|^2-z_1barz_2-z_2bar z_1+|z|^2`
Here `(x-x_0)^2+(y-y_0)^2`
and `(x-x_0)^2+(y-y_0)^2=4r^2`
since `alpha and 1/alpha` lies on first and second respectively
`therefore " "|alpha-z_0|^2=r^2 and |(1)/(bar alpha -z_0)|^2=4r^2`
`rArr (alpha-z_0)(baralpha-alpha _0)=r^2`
`rArr |alpha^2|-z_0 bar alpha- bar z_0 alpha+|z_0|^2=r^3`
and `|1/(alpha)-z_0|^2=4r^2`
`rArr 1/alpha-z_0=4r^2`
`rArr 1/(|alpha|^2)-(z)/alpha-(barz_0)/(baralpha)+|z_0|^2=4r^2`
Since `|alpha|=alpha.alpha`
`1/(|alpha|^2)-(z_0.baralpha)/(|alpha|^2)-barz_0/(|alpha|^2).alpha+|z_0|^2=4r^2`
`rArr 1-z_0baralpha-barz_0alpha+|alpha|^2|z_0|^2=4r^2|alpha|^2`
On subtracting Eqs. (i) and (ii) ,we get
`(|alpha|^2-1)(1-|z_0|^2)=r^2(1-4|alpha|^2)`
`rArr(|alpha|^2-1)(1-|z_0|^2)=r^2(1-4|alpha|^2)`
`rArr (|alpha|^2-1)(1-(r^2+2)/(2))=r^2(1-4|alpha|^2) `
Given `|z_0|^2=(r^2+2)/(2)`
`rArr (|alpha|^2-1).((-r^2)/2)=r^2(1-4|alpha|^2)`
`rArr |alpha|^2-1=-2+8|alpha|^2`
`rArr |alpha|^2-1=-2+8|alpha|^2`
`rArr " "7|alpha|^2=1`
`therefore |alpha|=1 //sqrt(2)`