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I have two circles as: $C_1: (x-x_1)^2+(y-y_1)^2=r_1^2$ and $C_2: (x-x_2)^2+(y-y_2)^2 =r_2^2$ and these circles have non-empty intersection.

In other words $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\leq r_1+r_2$.

Now I define a third circle as:

$C: (x-x_0)^2+(y-y_0)^2 = r_0^2$ where

$x_0=x_1(1-t)+x_2t$

$y_0=y_1(1-t)+y_2t$

$r_0=\sqrt{x_0^2+y_0^2-(x_1^2+y_1^2-r_1^2)(1-t)-(x_2^2+y_2^2-r_2^2)t}$ where $0\leq t \leq1$

Claim: C contains the intersection of $C_1$ and $C_2$ for all values of t such that $0 \leq t \leq 1$.

How can i prove this claim?

  • 0
    Your $r$ term seems a little weird. Two of the summands in there are $0$. And do you mean the centers of the circles are $(x_1,\ y_1)$ and $(x_2,\ y_2)$? As you have it, the two circles are centered at the origin and $x_i$ and $y_i$ are variables.2012-10-04
  • 0
    You didn't address the issues I mentioned. It makes no difference if you're talking about circles or disks. Do you mean for the equation to hold for every $x_1,\ y_1,\ x_2,\ y_2$ satisfying the inequalities?2012-10-04
  • 0
    just check now, sorry ;)2012-10-04
  • 0
    To simplify the computation, you can, without loss of generality, assume that $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(1,0)$.2013-03-31

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