 # prove that every circle passing through points -1 and +1 have an equation of the form zz'+ikz-ikz'-1=0 where k is real and z' = z conjugate

11 years ago

Dear Rachneet,

There are two approaches to this problem:

Approach 1: Using co-ordinate geomety

Family of circles using diametric form with points (-1,0) and (1,0) => x2+y2=1as diameter and line passing through these two points =>y=0 will be

x2+y2-1 +py=0 where p is a variable

=> |z|2-1+p(z-z')/2i=0

=> zz'+ikz-ikz'-1=0 where k=-p/2 is real

Approach 2:

Equation of circle: Arg[(z-1)/(z+1)]=θ

=> (z-1)/(z'-1)=[(z+1)/(z'+1)]ei2θ

=> Simplify to get (zz'-1)+(z'-z)[(ei2θ+1)/(ei2θ-1)]=0

=> (zz'-1)+(z'-z)(-icotθ)=0 where (ei2θ+1)/(ei2θ-1)=-icotθ

=> zz'+ikz-ikz'-1=0 where k=cotθ is real

Hope this helps.

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