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rachneet kaur Grade:
        

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

6 years ago

Answers : (1)

AskiitiansExpert Mohit-IITD
21 Points
										

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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6 years ago
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