Solution:
Obviously, the segment $BC$ is tangent to the circle $c$. Let $M$ be the point of tangency (Figure 6).

Call $Q$, $M$ the tangency points of $BA$, $BC$ with $c^{\prime}$ and $c$ respectively, and call $H$ the midpoint of segment $AC$. It is well known that
$$AQ = \frac{1}{2}(AO_{1} + AB - BO_{1}) \text{ and } CM = \frac{1}{2}(BO_{1} + BC - CO_{1})$$
and so
$$AQ + CM = \frac{1}{2}(2BC - AC) = a - \frac{1}{2}AC = a - HC$$
The triangles $KAQ$ and $OCM$ are similar and this implies
$$\begin{aligned} \frac{KQ}{OM} & = \frac{AQ}{CM} \Leftrightarrow \frac{KQ}{AQ} = \frac{OM}{CM} \Leftrightarrow \frac{r}{AQ} = \frac{R}{CM} = \frac{R + r}{AQ + CM} = \frac{R + r}{a - HC} \Leftrightarrow \\ \frac{r}{AQ} & = \frac{R + r}{a - HA} \end{aligned}$$
If $AZ$ is the bisector segment of triangle $BAH$ it holds
$$\angle AZH = 90^{\circ} = \frac{1}{2} \angle BAC \text{ and } \angle KAQ = \frac{1}{2}(180^{\circ} - \angle BAC) = 90^{\circ} - \frac{1}{2} \angle BAC$$
Therefore, from the similar triangles $KQA$ and $AHZ$ we get
$$\frac{-AH}{ZH} = \frac{r}{AQ} \stackrel{(6)}{=} \frac{R + r}{a - HA}$$
Also, from the bisector theorem in triangle $ABH$ it holds
$$ZH = \frac{AH \cdot BH}{a + AH}$$
and from (7) it follows
$$\frac{R + r}{a - HA} = \frac{AH}{\frac{AH \cdot BH}{a + AH}} \Rightarrow R + r = \frac{a^{2} - AH^{2}}{BH} = \frac{BH^{2}}{BH} = BH$$
So
$$HA^{2} = a^{2} - (R + r)^{2} \Leftrightarrow HA = \sqrt{a^{2} - (R + r)^{2}}$$
and finally the area of triangle $ABC$ in terms of $a$, $R$, $r$ is:
$$(ABC) = AH \cdot BH = (R + r) \sqrt{a^{2} - (R + r)^{2}}$$