OA = 3，

∴AB=5

So OB=AB2? OA2=4，

Therefore, the coordinates of point B are: (0,4),

Let the analytical formula of straight line AB be: y=kx+b(k≠0),

Replace it with a (-3,0) and b (0 0,4)? 3k+b=0b=4，

Solution: k = 43b = 4,

∴AB straight line analytical formula: y = 43x+4.

Substituting a (-3,0), C (- 1, 0) and B (0 0,4) into the parabolic analytical formula, we can get: 9a? 3b+c=0a？ b+c=0c=4，

Solution: a = 43b = 163c = 4,

Therefore, the analytical formula of parabola is y = 43x2+ 163x+4.

(2) Let the coordinates of P(x, 43x+4) and d be known as: (2,0).

① If △ABO∽△APD,

Then AOAD=ABAP=BOPD, which means 35=4DP.

Solution: DP=203,

Therefore, the coordinate of point P is (2,203).

② If △ABO∽△ADP,

ABAD=AOAP, which means 55=3AP,

Solution: AP=3,

Then (x+3)2+(43x+4)2=32,

Solution: x 1=-65, x2=-245.

Therefore, the coordinates of point P are: (-65, 125).

(3) the radius ⊙ d is r=2,

When the coordinate of point P is (2,203), the radius ⊙A is AP=253, and AD = 5 < 253-2.

Therefore, at this time, two circles contain;

When the coordinates of point P are: (-65, 125), the radius ⊙A is AP=3, and AD=5=3+2.

Therefore, the two circles are circumscribed at this time.