Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days


Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number


Mon to Sat - 11 AM to 8 PM

CBSE - XII Science - Mathematics

Complex nos

Asked by 10th April 2010, 1:51 PM
Answered by Expert


Dear Student,


Let the center of the circle which touches both circles |z-z1|=a and |z-z2|=b be z0.Let the radius of this circle be r. So the equation of this circle becomes |z-z0|=r.

Now as this circle touches |z-z1|=a, therefore z1 is at a distance of (r+a) from z0.

=> |z0-z1|=a+r               (1)

Similarly, as this circle also touches |z-z2|=b, therefore z2 is at a distance of (r+b) from z0.

=> |z0-z2|=b+r               (2)

Subtracting (2) from (1), we get

|z0-z1| - |z0-z2| = a - b

Since z0 is a variable point, replace it by z and replace (a-b) by k.

=> |z-z1| - |z-z2| = k

This is a standard equation for a hyperbola. therefore, the locus of the centre of the circle which touches the given two circles is a hyperbola.


So, the correct answer is option (c).


Regards Topperlearning.

Answered by Expert 28th April 2010, 3:26 PM

Rate this answer

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Report an issue
Your answer has been posted successfully!