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.