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?