Let the complex numbers ( ) and ( ( 1 ) ) lie on circles ( (x-x₀ )^2+ (y-y₀ )^2=r^2 ) and ( (x-x₀ )^2+ (y-y₀ )^2=4 r^2 ) respectively. If (z₀=x₀+i y₀ ) satisfies the equation (2 |z₀ |^2= ) (r^2+2 ), then (| |= )

Step-by-step solution

As point ( ) lies on the circle ( aligned & (x-x₀ )^2+ (y-y₀ )^2 =r^2 & | -z₀ |^2=r^2, where z₀ =x₀+i y₀ & | |^2+ |z₀ |^2- ( z ₀+ z₀ ) =r^2 (i) aligned ) ( 1 ) lies on the circle ( (x-x₀ )^2+ (y-y₀ )^2=4 r^2 ) ( | 1 -z₀ |^2=4 r^2 ) ( aligned & 1 | |^2 + |z₀ |^2- ( z ₀ | |^2 + z₀ | |^2 )=4 r^2 & 1+ |z₀ |^2| |^2- ( z ₀+ z₀ )=4 r^2| |^2 (ii) aligned ) By subtracting Eqs. (i) and (ii), we get ( array cc & 1-| |^2- |z₀ |^2 (1-| |^2 )=r^2 (4| |^2-1 ) & (| |^2-1 ) ( |z₀ |^2-1 )=r^2 (4| |^2-1 ) & |z₀ |^2= r^2+2 2 , we get