Intersection of Two Circles

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?

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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. – EuYu Oct 4 '12 at 18:53
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? – EuYu Oct 4 '12 at 19:10
just check now, sorry ;) – kotoll Oct 4 '12 at 19:11
To simplify the computation, you can, without loss of generality, assume that $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(1,0)$. – Aline Parreau Mar 31 '13 at 19:13

If point $(x_3,y_3)$ is in the intersection of $C_1$ and $C_2$ then $(x_3-x_1)^2+(y_3-y_1)^2=x_3^2+x_1^2-2x_3x_1+y_3^2+y_1^2-2y_3y_1 \leq r_1^2$ and $(x_3-x_2)^2+(y_3-y_2)^2=x_3^2+x_2^2-2x_3x_2+y_3^2+y_2^2-2y_3y_2\leq r_2^2$
Then the distance from $(x_0,y_0)$ is $\sqrt{(x_3-(1-t)x_1-tx_2)^2+(y_3-(1-t)y_1-ty_2)^2}=\sqrt{x_3^2+t^2x_2^2+(1-t)^2x_1^2-2(1-t)x_3x_1-2tx_3x_2+t(1-t)x_1x_2+y_3^2+t^2y_2^2+(1-t)^2y_1^2-2(1-t)y_3y_1-2ty_3y_2+t(1-t)y_1y_2}$ .
$\leq \sqrt{x_1^2(1-t)^2+x_2^2t^2+2x_1x_2t(1-t)+y_1^2(1-t)^2+y_2^2t^2+2y_1y_2t(1-t)-x_1^2-y_1^2+r_1^2+tx_1^2+ty_1^2-tr_1^2-tx_2^2-ty_2^2+tr_2^2}$
$\leq \sqrt{x_0^2+y_0^2-x_1^2-y_1^2+r_1^2+tx_1^2+ty_1^2-tr_1^2-tx_2^2-ty_2^2+tr_2^2}$
$= \sqrt{x_0^2+y_0^2-(x_1^2+y_1^2-r_1^2)(1-t)-(x_2^2+y_2^2-r_2^2)t}=r_0$