• Select a point on the bigger circle and work with midpoints.
• Using the theorem which states in a triangle $ABC$, $M$ and $N$ are midpoints of $AB$ and $AC$ if and only if   $MN=\frac{1}{2}BC$ and $MN||BC$

Figure 01: dividing $r_1$ equally

Figure 02: Construction

In Figure 1:

We first draw $AB=r_1$ and draw it's midpoint.

$M$. so $AM=\frac{r_1}{2}$

Now in Figure 2:

We do construction. First we select a random point E on biggest circle.

So $AE=r_3$.

Now we draw it's midpoint $F$.

Now we draw a circle with center $F$ with radius $\frac{r_1}{2}$.Name the circle $w$.

We know, $r_1+r_3 \geq 2r_2$

So, $\frac{r_1}{2}+\frac{r_3}{2} \geq r_2$

So, $w$ must intersect the 2nd big circle.Let's say at $G$.

Now draw a line parallel  to $FG$ through $A$

Which intersects small circle at $H$.

Now $AF=FE$,$FG || AH$ and $AH=2FG$

So, $HG=GE$ and $H,G,E$ collinear.

Thus we do the construction.