Question

Let complex numbers  alpha  and 1/ baralpha
lie on circles  (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 + i y_0  satisfies the equation 2|z_0|^2 = r^2 + 2, then | alpha | = ?

(a) 1/3

(b) 1/sqrt2

(c) 1/2

(d) 1/sqrt7

Solution

|alpha-x_0-iy_0|=r

|alpha-Z_0|=r

|1/baralpha-x_0-iy_0|=2r

|1/baralpha-Z_0|=2r

Given that, 2|Z_0|^2=r^2+2

We know that ZbarZ= |Z|^2

=>(alpha-Z_0)(baralpha-barZ_0)=r^2

=>alphabarZ_0+baralphaZ_0=|alpha|^2+1-r^2/2

=>(1/baralpha-Z_0)(1/alpha-barZ_0)=4r^2

=>(1-baralphaZ_0)(1-alphabarZ_0)=4baralphaalphar^2

=>-alphabarZ_0-baralphaZ_0=4|alpha|^2r^2-1-|alpha|^2(r^2/2+1)

=>-alphabarZ_0-baralphaZ_0=7|alpha|^2r^2/2-1-|alpha|^2

-|alpha|^2-1+r^2/2=7|alpha|^2r^2/2-1-|alpha|^2

|alpha|=1/sqrt(7)

Therefore, option (d) is correct.