Correct Answer - Centre = `(alpha - k^(2) beta)/(1 - k^(2))` , Radius = `|(k (alpha - beta))/(1-k^(2))|`
As know we know `|z|^2=z.barz`
Given `(|z-alpha|^2)/(|z-beta|^2)=k^2`
`(z-alpha)(barz-baralpha)=k^2(barz-barbeta)`
`rArr |z|^2-alphabarz-baralphaz+|alpha^2|=k^2(|z|^ 2-betabarz-barbetaz+|beta|^2)`
` rArr |z|^2(1-k^2)-(alpha - k^2beta)barz -(baralpha- betak^2)z`
`rArr |z|^2-((alpha-k^2beta))/((1-k^2))z-((baralpha-barbetak^2))/((1-k^2))z+(|alpha|^2-k^2|beta|^2)/((1-k^2))=0...(i)`
On comparing with equation of circle ,`|z|^2+a barz+barz+0`
Whose center is (-a) and radius `=sqrt(|a|^2-b)`
`therefore` Center for Eq. (i) `=(alpha-k^2beta)/(1-k^2)`
and radius `=sqrt(((alpha-k^2beta)/(1-k^2))((baralpha-k^2beta)/(1-k^2))-(alphabaralpha-k^2betabarbeta)/(1- k^2))`
`=|(k(alpha-beta))/(1- k^2)|`