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Message from discussion Intersection of two disks

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Newsgroups: sci.math
Date: Sun, 7 Oct 2012 09:16:37 -0700 (PDT)
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Subject: Re: Intersection of two disks
From: Narasimham <mathm...@gmail.com>
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On Saturday, October 6, 2012 2:17:55 AM UTC+5:30, kamuran turksoy wrote:
> I have two disks:
> C1: (x-a1)^2+(y-b1)^2<= r1^2
> C2: (x-a2)^2+(y-b2)^2<= r2^2
> These disks have non-empty intersection.
> I define the third circle as:
> C3: (x-a3)^2+(y-b3)^2<=r3^2 where
> a3=a1*(1-t)+a2*t
> b3=b1*(1-t)+b2*t
> r3=sqrt(a3^2+b3^2-(a1^2+b1^2-r1^2)*(1-t)-(a2^2+b2^2-r2^2)*t)
> where 0<=t<=1.
> Claim: C3 contains the intersection of C1 and C2 for all values of t such that  0 <= t <= 1.
> Numerically when I substitute t values and check it, the claim works. However I could not prove it. Any suggestions?

Hint : The Power (product of tangent lengths) is negative originating from a point contained inside the circle (and positive if outside).

Narasimham

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