InterviewSolution
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InterviewSolution
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Find the centre and radius of the circle formed by all thepoints represented by `z = x + iy` satisfying the relation `|(z-alpha)/(z-beta)|= k (k !=1)`, where `alpha and beta` are the constant complex numbers given by `alpha = alpha_1 + ialpha_2, beta = beta_1 + ibeta_2`. |
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Answer» 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)|` |
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