Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        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`
7 years ago

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.
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 2,900 off
USE CODE: CHENNA
Get extra Rs. 2,332 off
USE CODE: CHENNA