Rolling circle on an ellipse

Take the ellipse to be the one given by $\frac {x^2}{16} + \frac {y^2}9 = 1.$ Let $(t,u)$ be the point of tangency, and assume without loss of generality that $(t,u)$ lies in the first quadrant. The tangent line at $(t,u)$ is $y - u = -\frac {9t}{16u}(x - t),$ so the radial line of the circle through $(t,u)$ is $y - u = \frac {16u}{9t}(x - t).$ Let $(c,d)$ denote the center of the circle. Then $(c,d)$ lies on the radial line at a distance of 1 from $(t,u)$ , which is to say $(c,d) = \left( t + \cos \theta , u + \sin \theta \right), \quad \theta = \arctan \frac {16u}{9t}.$ The line joining the centers of the ellipse and circle is $y = \frac dc x,$ and the distance between the point of tangency $(t,u)$ and this line is measured along the perpendicular line $y - u = -\frac cd (x - t).$ These two perpendicular lines intersect at $\left( \frac {c(ct + du)}{c^2 + d^2} , \frac {d(ct + du)}{c^2 + d^2}, \right).$ The problem is to maximize the distance between this point and $(t,u)$ . Let $r$ denote this distance: $r = \sqrt{\left( t - \frac {c(ct + du)}{c^2 + d^2} \right)^2 + \left( u - \frac {d(ct + du)}{c^2 + d^2}\right)^2}.$ Notice that $c$ and $d$ are given in terms of $t$ and $u$ , and $u$ can be expressed in terms of $t$ as $u = 3\sqrt{1 - \frac {t^2}{16}}.$ This means $r$ can be regarded as a function of $t$ . The graph of $y = r(t)$ , with $0 \leq t \leq 4$ , is shown below. The maximum value of $r$ on this interval is approximately $0.21988$ , so the desired value is $\lfloor 10000 \times 0.21988 \rfloor = \boxed{2198}$ .

Incidentally, with a couple sign changes, one can solve the problem when the circle is rolling in the interior of the ellipse. In this case, the maximum distance is approximately $0.38267$ . Maybe the wording of the problem should be changed to specify that the circle rolls on the exterior of the given ellipse .