Let $B_1$ be the symmetrical of the point $B$ with respect to the line $IC$, and let $C_1$ be the symmetrical of the point $C$ with respect to the line $IB$. Denote by $B_2$ the midpoint of the segment $IB_1$, and by $C_2$ the midpoint of the segment $IC_1$. The circles $\gamma'$ and $\delta'$ have diameters $IB_1$ and $IC_1$, hence their centers are $B_2$ and $C_2$. The line passing through the meeting points of circles $\gamma'$ and $\delta'$ is perpendicular to the centers line. Since $I$ belongs to both circles $\gamma'$ and $\delta'$, the requirement comes to $OI \perp B_2C_2$. Since $B_2C_2$ is midline in triangle $IB_1C_1$, one has $B_1C_1 \parallel B_2C_2$, and the requirement writes $OI \perp B_1C_1$. It is enough to show $OB_1^2 - OC_1^2 = IB_1^2 - IC_1^2$. Let $D$ be the projection of $I$ onto the line $BC$. $D$ is the touching point of the incircle of triangle $ABC$ with the side $BC$, and $BD = p - b$, $DC = p - c$. From symmetry considerations we have $IB_1 = IB$, $IC_1 = IC$, and
$$\begin{aligned} IB_1^2 - IC_1^2 &= IB^2 - IC^2 = \\ &= DB^2 - DC^2 = \\ &= (p-b)^2 - (p-c)^2 = \\ &= a(c-b). \end{aligned}$$
Denote by $M$ the midpoint of the side $AC$, and by $N$ the midpoint of the side $AB$; then we have
$$\begin{aligned} OB_1^2 - OC_1^2 &= \\ &= (OM^2 + MB_1^2) - (ON^2 + NC_1^2) = \\ &= (MB_1^2 + OA^2 - MA^2) - \\ &\qquad (NC_1^2 + OB^2 - NB^2) = \\ &= (MB_1^2 - MA^2) - (NC_1^2 - NB^2), \end{aligned}$$
since $OA = OB$. As $MB_1^2 - MA^2 = MB_1^2 - MC^2 = (MB_1 - MC)(MB_1 + MC) = a(a - b)$, and analogously $NC_1^2 - NB^2 = a(a - c)$, it follows that $OB_1^2 - OC_1^2 = a(c - b)$, which is what was left to prove.