Avoid the Annoying Circular Objects

Since the radius of a circle is perpendicular to the tangent line, we have $\angle BDE =\angle ACE = 90^{\circ}$ . Also $\angle BED = \angle AEC$ (since they are vertical angles), and $\overline{AC}=9=\overline{BD}$ . Hence $\Delta AEC$ and $\Delta BED$ are congruent by AAS similarity. Since the centers of the circles are located at $(9,9)$ and $(45,9)$ , we have $\overline{EB} = \frac{1}{2} (45-9) = 18$ . So, by the Pythagorean theorem, $\overline{ED} = \sqrt{18^2-9^2} = 9\sqrt{3}.$

Note $\angle EBD = \arccos \frac{9}{18} = 60^{\circ}$ , and $\angle DBG = 90-\angle EBD = 30^{\circ}$ . So $\overset{\frown}{AD} = \frac{2\pi (9)}{12} = \frac{9\pi}{6}$ . Finally, observe that $\overline{GH} = 9$ . By symmetry, the path from $O$ to $H$ has length $2 \left(9+9\sqrt{3} + \frac{9\pi}{6}\right) = 18 + 18\sqrt{3} +3\pi.$ Therefore $a+b+c+d = 42$ .

Let $(a,b)$ and $(p,q)$ be the two points on the first & the second circle where the common internal tangent intersects the circles.

It is clear that (9,9) & (45,9) are the centers of the circles both of radius = 9. Now we may write the following equations:

$(a-9)^2+(b-9)^2=81$ and $(p-45)^2+(q-9)^2=81$

Then:
$9=27(q-b)/(p-a) + c$ (c = intercept of the tangent on the y-axis), so:
$(q-b)/(p-a)=-(9-a)/(9-b),$
$(9-b)/(9-a)=(q-9)/(p-45).$

Solving all of these, we find $a,b,c,p,q = \frac{27}{2},9-\frac{9}{2}\sqrt{3},9-9\sqrt{3},\frac{81}{2},\frac{9(2+\sqrt{3})}{2}.$

The path is (0,0)->(9,0)->(a,b)->(p,q)->(45,18)->(54,18) which can now be calculated as $18+18\sqrt 3 + 3\pi$ , so $a+b+c+d = 42.$