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.