Luckily, I have done this problem. Wait a minute. 1 Because AB is the diameter g of a circle, ∠ AOB = 90, so the origin O is always on ⊙ G. 2 It is easy to prove that △ABC is a right triangle, and AB/AC=6/3=2, so ∠ B = 30 crosses point C to make CD perpendicular to the X axis and connect OC. Because the circumferential angles of the same arc are equal, ∠ AOC = ∠ B = 30, so x/y=OD/CD=√3, so Y! Here, the movement of point C has two stages. When BC is perpendicular to the Y axis, the distance of C is the farthest, which is set to C 1, and then it gets closer and closer. When B moves to the origin, set C2 total distance = (oc1-oc)+(oc1-oc2) = (6-3)+(6-3 √ 3) = 9-3 √ 3 There are no pictures, so it is difficult to understand. Let me draw a picture for you. Wait a minute. . Sorry, the attachment in the program folder was deleted, and the drawing tool could not be found. . . . . Find it yourself. Sorry. . . . I found it, hehe, it's finished, it's not good-looking, please forgive me.

V= 1 I'm exhausted, sweat! !