2021 Fall AMC 12B Problem 22

Place $C = (0, 0),$ $B = (6, 0),$ and $A = (0, 8),$ so the right angle is at $C.$ Circle $O$ is tangent to line $BC$ (the $x$-axis) at $B,$ so $O = (6, k).$ Setting $OA = OB$ gives $36 + (k - 8)^2 = k^2,$ so $k = \tfrac{25}{4}$ and $O = \left(6, \tfrac{25}{4}\right).$ Circle $P$ is tangent to line $AC$ (the $y$-axis) at $A,$ so $P = (h, 8).$ Setting $PB = PA$ gives $(h - 6)^2 + 64 = h^2,$ so $h = \tfrac{25}{3}$ and $P = \left(\tfrac{25}{3}, 8\right).$ Then $OP = \sqrt{\left(\tfrac73\right)^2 + \left(\tfrac74\right)^2}$ $= 7\sqrt{\tfrac{25}{144}}$ $= \dfrac{35}{12}.$ Thus, the correct answer is C.