Author Topic: Circles of the Complex Form |z-a|=k|z-b| (Read 2592 times) Tweet Share

Jayz2398 · « **on:** September 22, 2016, 01:39:17 pm »

Hey all!

I've been doing some complex no. revision and looking over some of the notes from the ATARNotes spesh lecture, and I'm hoping to get some thoughts on this complex form which I've been told is a circle:
$|z-a|=k|z-b|$

Doing some algebra, I concluded that this form holds true to give a circle as long as $k\neq 0 \cup 1$ .

Do you guys agree? I'd really appreciate your feedback if you've worked with this kind of form before. I remember seeing it in my first Spesh SAC this year, but never looked it up until now.

Thanks,
Jack

RuiAce · « **Reply #1 on:** September 22, 2016, 01:42:04 pm »

Quote from: Jayz2398 on September 22, 2016, 01:39:17 pm

Hey all!

I've been doing some complex no. revision and looking over some of the notes from the ATARNotes spesh lecture, and I'm hoping to get some thoughts on this complex form which I've been told is a circle:
$|z-a|=k|z-b|$

Doing some algebra, I concluded that this form holds true to give a circle as long as $k\neq 0 \cup 1$ .

Do you guys agree? I'd really appreciate your feedback if you've worked with this kind of form before. I remember seeing it in my first Spesh SAC this year, but never looked it up until now.

Thanks,
Jack

$\text{Yes.}$
$\text{To explicitly prove this result (geometrically) requires minimum second year university maths.}\\ \text{The basic idea is that for the following special cases:}$
$k=1\\ \text{The distance to }a\text{ is the SAME as that to }b\\ \text{Because the distances are strictly equal, no curvature occurs.}\\ \text{This can be seen as a 'degenerate case'.}$
$\text{Effectively, if the distance is the same then we get a LINE.}\\ \text{We get the perpendicular bisector of the interval joining }a\text{ and }b$
$k=0\\ \text{This immediately causes the RHS to vanish.}\\ \text{Since the RHS is now 0, you have the single point }a\\ \text{which is your locus.}$
$\text{Remark: Of course, }k\text{ must be non-negative or we have a contradiction.}$

Jayz2398 · « **Reply #2 on:** September 26, 2016, 12:43:54 pm »

Thank you very much!

I'd had a muck around with the algebra and found that it definitely didn't correspond to a circle for k = 1 or 0, but I didn't know about negative values of k.

Search the forums now!

We have moved!

Author Topic: Circles of the Complex Form |z-a|=k|z-b| (Read 2592 times) Tweet Share

Jayz2398

Circles of the Complex Form |z-a|=k|z-b|

RuiAce

Re: Circles of the Complex Form |z-a|=k|z-b|

Jayz2398

Re: Circles of the Complex Form |z-a|=k|z-b|

Recent Posts

User:
Password: