$ H$ is the orthocenter. $ A', B', C'$ are not feet of perpendiculars from $ A, B, C$ to $ BC, CA, AB,$ but points on the A, B-, C-altitudes $ AH, BH, CH,$ respectively. (a) Assuming $ \triangle A'B'C' \sim \triangle ABC$ are oppositely similar, otherwise $ AA' = BB' = CC' = x = 0$ satisfies the conditions. $ \triangle A'B'C' \sim \triangle ABC$ $ \Longrightarrow$ $ A'B'C'H$ is cyclic. Assume $ AA' = BB'.$ Locus of circumcenters $ O'_c$ of the $ \triangle B'HA'$ is a line $ p_c \parallel CI$ parallel to the internal bisector of the $ \angle BCA.$ Likewise, assuming $ BB' = CC',$ locus of circumcenters $ O'_a$ of the $ \triangle C'HB'$ is a line $ p_a \parallel AI$ and assuming $ CC' = AA',$ locus of circumcenters $ O'_b$ of the $ \triangle A'HC'$ is a line $ p_b \parallel BI.$ Let $ P_a, P_b, P_c$ be reflections of the circumcenter $ O$ in $ BC, CA, AB.$ Then $ P_a \in p_a, P_b \in p_b, P_c \in p_c.$ $ \triangle P_aP_bP_c \cong \triangle ABC$ are centrally congruent with similarity center $ N$ (common 9-point circle center). Lines $ p_a, p_b, p_c$ are internal bisetors of the $ \triangle P_aP_bP_c,$ concurrent at its incenter $ F.$ Circle $ (F)$ with radius $ FH$ is the only circle through $ H$ cutting $ AH, BH, CH$ again at $ A', B', C',$ such that $ AA' = BB' = CC' = x > 0.$ The Fuhrmann circle of $ \triangle ABC$ goes through $ H$ and cuts $ AH, BH, CH$ again at $ A', B', C',$ such that $ AA' = BB' = CC' = 2r$ (see the last reply at http://www.mathlinks.ro/Forum/viewtopic.php?t=101687) $ \Longrightarrow$ $ (F)$ is the Fuhrmann circle and $ x = 2r.$ (b) Let $ X, Y, Z$ be reflections of $ A', B', C'$ in the internal angle bisectors $ AI, BI, CI.$ $ AX = BY = CZ = x$ $ \Longrightarrow$ $ X \in AO, Y \in BO, Z \in CO$ are on a circle $ \mathcal P(O, x - R)$ (radius can be positive or negative). Let $ k \equiv A'B'C'$ be the collinarity line and $ k_x, k_y, k_z$ reflections of $ k$ in $ AI, BI, CI.$ Let $ k_x, k_y$ intersect at $ Q$ (different from $ X, Y$). By angle chase for the $ \triangle B'HA'$ and quadrilateral $ XOYQ,$ $ \angle B'HA' = \angle XQY = \frac {\angle XOY}{2} \mod \pi$ $ \Longrightarrow$ $ Q \in \mathcal P(O, x - R).$ Similarly, $ k_y, k_z$ intersect at a point on the circle $ \mathcal P$ different from $ Y, Z$ $ \Longrightarrow$ $ k_x, k_y, k_z$ concur at $ Q \in \mathcal P.$ Let $ U, V, W$ be reflections of $ Q$ in $ AI, BI, CI.$ The incenter $ I$ is the circumcenter of the cyclic quadrilateral $ QUVW,$ because perpendicular bisectors $ AI, BI, CI$ of its 3 sides $ QU, QV, QW$ concur at $ I.$ $ U, V, W \in k$ are collinear $ \Longrightarrow$ this is possible only when $ I \equiv Q$ ($ \equiv U \equiv V \equiv W$), therefore $ I \in \mathcal P,$ $ x - R = \pm OI$ and $ AA' = BB' = CC' = x = AX = BY = CZ = R \pm OI.$

We can solve part a easier . We know $A'B'C'H$ in cycle . so $ A'H . B'C' + HB' . A'C' = HC' . A'B' $ $(x-AH) . B'C' +(x-BH) . A'C' = (CH - x) . A'B' $ $\triangle A'B'C' \sim \triangle ABC$ so $ AB = k A'B' $ and .... so $(x-AH) . BC +(x-BH) . AC = (CH - x) . AB $ so $ 2xp = BC.AH + AC.BH + AB.CH = 4S = 4rp $ so $x=2r$ I mean $2p = AB + BC + CA $ and $S$ is the area of the triangle .

Is there a strict proof for $A'B'C'H$ being cyclic? Angle chasing might work only we presume a specific configuration of those points.