∴C(0,6),D( 1,8)，

Let a straight line CD: y = kx+b (k ≠ 0) be substituted into c and d to get b = 68 = k+6.

The solution is k = 2b = 6,

∴CD linear analytical formula: y=2x+6, when y=0, x=-3.

∴e(-3,0)；

(2) Let y=0 get -2x2+4x+6=0,

The solution is x 1=- 1, x2=3,

∴A(- 1,0),B(3,0)，

Also: o (0 0,0), E (-3,0),

∴ center o/kloc-0 with OE as the diameter/(? 32,0), radius r 1 = 32.

Let P(t, 2t+6),

(t+32) 2+(2t+6) 2 = 32 from PO 1 = 32,

Solution: t 1=- 125, t 2=-3 (shed),

∴P(？ 125,65),

∴PA=855,AO 1= 12，

DC = 5，CB = 35，DB = 2 17，

∴dcao 1=cbpo 1=dbpa=25，

∴△bcd∽△po 1a；

(3)①O 1(？ 32，0)，r 1=32，O2(0，m)

According to the meaning of the question, it is obvious that point O2 is below point C r2=O2C=6-m,

When ≥O2 and ≥o 1 are circumscribed, O 1O2=r 1+r2,

Substitute (32) 2+m2 = 32+(6? m)，

Solution: m 1 = 185, m2 = 2 (s),

When ⊙O2 and ⊙O 1 are inscribed, O 1O2=|r 1-r2|,

Substitute (32) 2+m2 = | 32? (6? m)|，

Solution: m 1 = 2, m2 = 185 (s),

∴m 1= 185,m2=2，

② O3 (0， 185)，O3 (0，2)，

When ⊙O 1 and ⊙O2 are circumscribed, O3 (32,0), O3 (0, 14 15), O3 (3212,0), O3 (0,0 107);

When ⊙O 1 and ⊙O2 are connected, O3 (? 45 14,0).