Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Intersection of two Disks
Date: Fri, 5 Oct 2012 22:24:09 +0000 (UTC)
Organization: Xoran Technologies
Lines: 22
Message-ID: <k4nmm9$si9$1@newscl01ah.mathworks.com>
References: <k4ng64$647$1@newscl01ah.mathworks.com>
Reply-To: <HIDDEN>
NNTP-Posting-Host: www-03-blr.mathworks.com
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
X-Trace: newscl01ah.mathworks.com 1349475849 29257 172.30.248.48 (5 Oct 2012 22:24:09 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Fri, 5 Oct 2012 22:24:09 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 1440443
Xref: news.mathworks.com comp.soft-sys.matlab:779984

"kamuran turksoy" <kamuranturksoy@gmail.com> wrote in message <k4ng64$647$1@newscl01ah.mathworks.com>...
>
> Claim: C3 contains the intersection of C1 and C3 for all values of t such that 0<=t<=1
=================

I assume you mean "C3 contains the intersection of C1 and C2"

Fix any x in the intersection of C1 and C2 and let p=[a2, b2]. Also, assume without loss of generality that  a1=b1=0. You can always translate space so that one of the circle centers is at the origin without changing the geometry of the intersection.

Under these assumptions, your formula for r3^2 reduces to

r3^2 = |p|^2*t^2+ (r1^2-r2^2-|p|^2)*t+r2^2

and the distance of x from [a3,b3] is

 |x-t*p|^2= |p|^2*t^2  - 2*dot(x,p)*t+|x|^2

We want to show that the r3^2>=  |x-t*p|^2 in the interval 0<=t<=1. Considering the above formulas, it is therefore necessary to show that

(r1^2-r2^2-|p|^2)*t+r2^2  - ( - 2*dot(x,p)*t+|x|^2)  >=0

is non-negative throughout 0<=t<=1. But the LHS of this equation is linear in t, so it is a simple matter to analyze where it is non-negative.