Let $(I,r)$ and $(I_a,r_a)$ denote the incircle and A-excircle of $\triangle ABC.$ Let $\varrho_1$ and $\varrho_2$ denote the radii of $k_1,k_2.$ If $k_1,k_2$ touch $AC$ at $D,E,$ then it's well-known that $ID \perp AI$ and $I_aE \perp AI_a$ $(\star).$ Substituting $OA_1=R-\varrho_1,$ $OA_2=R+\varrho_2$ and ${A_1A_2}^2=DE^2+(\varrho_2-\varrho_1)^2$ into the required expression gives
$4R^2+(\varrho_2-\varrho_1)^2+4R(\varrho_2-\varrho_1)-DE^2-(\varrho_2-\varrho_1)^2=4R^2 \Longrightarrow$
$DE^2=4R \cdot (\varrho_2-\varrho_1).$
But $(\star)$ yields $\varrho_1=r \cdot \sec^2 \frac{A}{2} \ , \ \varrho_2=r_a \cdot \sec^2 \frac{A}{2} \ , \ DE=II_a \cdot \sec \frac{A}{2}$
$\Longrightarrow {II_a}^2 \cdot \sec^2 \frac{A}{2}=4R \cdot \sec^2 \frac{A}{2}(r_a-r) \Longrightarrow {II_a}^2=4R \cdot (r_a-r),$
which is a known triangle identity. Hence the proposed relation is proved.