mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?

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- June 18th 2006, 04:26 AMNaveshComplex No.
mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?

- June 18th 2006, 07:00 AMSoroban
Hello, Navesh!

It can be solved by "eyeballing" the problem if you know a particular theorem.

Quote:

, where represents a circle. .How?

The distance is proportional to the distance .

By definition, the locus of is a*Circle of Apollonius.* - June 18th 2006, 07:35 AMmalaygoelQuote:

Originally Posted by**Navesh**

Let

Let

Now,

This is equivalent to

It represents a circle if k is not equal to one. Do you know why? - June 18th 2006, 09:58 AMThePerfectHacker
malaygoel you used LaTex code I am proud of you ;)

Use,

Code:`\mbox{ "type you sentences here" }`

- June 18th 2006, 08:06 PMNaveshcircle of appoloniusQuote:

Originally Posted by**Navesh**

- June 18th 2006, 09:38 PMNaveshQuote:

Originally Posted by**Navesh**

- June 19th 2006, 06:52 AMNaveshQuote:

Originally Posted by**malaygoel**

also simplifying the eqn. we get

which makes the constt. term +ve, hence makes the circle real. Am I OK? - June 19th 2006, 08:36 AMSoroban
Hello, Navesh!

Quote:

also simplifying the eqn. we get

. . . not quite

which makes the constt. term +ve, hence makes the circle real. Am I OK?

We have: .

Expand: .

Rearrange: .

Factor: .

Hence: .

. . .Or: .

And we see the restrictions on the constants.

Obviously:

But equally important, the right side must be positive.

Hence: . and

. . . or: . and

- June 20th 2006, 09:01 AMNaveshQuote:

Originally Posted by**Soroban**