x2 = 4ay    ...........1
since circle touches the parabola at (6,9) so this point will satisfy the equation of both the curves
so , from eq 1
 36 = 36a
  a=1 , so eq 1 becomes
x2 = 4y            
now differentiating above eq       
2x = y1 = m (slope of tangent)
at (6,9) , m = 3        ..........2
let (h,k) be the center then slope of line passing through center of circle & the point of contact is given by
  m1 = (k-9)/(h-6)        .........3
mm1 = -1   or
(k-9)/(h-6) . 3 = -1        
 h + 3k = 33                      ................4
now focus of parabola is (0,1) ....
eq of circle is x2+y2 +2gx+2fy+c = 0                              (g=-h , f=-k)
 so eq of circle : x2+y2-2hx-2ky+c      
this circle passes through (0,1) & (6,9) so its eq will satisfy these points therefore
-2k+1+c = 0            ...................5
-12h-18k+117+c = 0             ....................6
from 5 & 6
3h+4k=29          ............7
solving 4 & 7 we get
(h,k) = (-9,14)
c = 27
now required eq of circle is
x2 + y2 18x - 28y + 27 = 0