Solution: (1) The center of the circle passing through ⊙O is OE⊥AC, and the vertical foot is E.

∴AE= 1/2AC= 1/2x, OE= under the root sign (AO? -AE？ ) = under the root sign (25- 1/4x? ).

∠∠deo =∠AOB = 90 ,∴∠d=90 -∠eod=∠aoe,∴△ode∽△aoe.

∴ OD/OE=AO/AE, ∫od = y+5, ∴ (y+5)/ under the root sign (25- 1/4x? )=5/(x/2)。

The ∴y's resolution function of x is: y = under root number 5 (100-x? )-5x/x。

The domains are: 0 < x < 5 and 2.

(2) When BD= 1/3OB, y=5/3, under the radical sign of 5 (100-x? )-5x/x。

∴x=6.

∴AE= 1/2x=3, OE= under the root sign (5? -3? )=4.

When the point O 1 is on the line OE, O 1E=OE-OO 1=2, O 1A= under the root sign (O 1E? +AE？ ) = under the root sign (2? +3? ) = 13 is under the root sign.

When the point O 1 is on the extension line of the straight line EO, O 1E=OE+OO 1=6, O 1A= under the root sign (O 1E? +AE？ ) = under the root sign (6? +3? ) = the root number 5 of 3.

The radius of ⊙o 1 is 13 or 35.

(3) Existence, when point C is the midpoint of AB, △ dcb ∽△ doc.

The proof is as follows: ∫ When point C is the midpoint of AB ∠ BOC = ∠ AOC = 1/2 ∠ AOB = 45,

oa = oc = ob，∴∠OCA =∠OCB =( 180-45)/2 = 67.5

∴∠DCB= 180 -∠OCA-∠OCB=45。

∴∠DCB=∠BOC .. and ∫∠d =∠d, ∴△DCB∽△DOC. ..

There is a point c that makes △ dcb ∴△ doc.