An ellipse, a line, and an inscribed circle

Label the diagram as follows:

Since the line has a slope of $2$ , $\frac{AB}{OB} = 2$ , and since $\triangle OAC \sim \triangle OBA$ by AA similarity, $\frac{AB}{OB} = \frac{AC}{OA}$ , or $2 = \frac{R}{OA}$ , which solves to $OA = \frac{1}{2}R$ .

By the Pythagorean Theorem on $\triangle OAC$ , $OC = \sqrt{R^2 + (\frac{1}{2}R)^2} = \frac{1}{2}\sqrt{5}R$ .

The circle then has an equation of $(x - \frac{1}{2}\sqrt{5}R)^2 + y^2 = R^2$ , so $y^2 = R^2 - (x - \frac{1}{2}\sqrt{5}R)^2$ .

Substituting this into the ellipse equation $\frac{x^2}{49} + \frac{y^2}{16} = 1$ and solving for $x$ gives $x = \frac{7}{66}(7\sqrt{5}R \pm 2\sqrt{53R^2 - 528})$ .

Since the circle is tangent to the ellipse, there should only be one $x$ value, so the discriminant $53R^2 - 528 = 0$ , and this solves to $R = 4\sqrt{\frac{33}{53}}$ .

Therefore, $\lfloor 1000R \rfloor = \lfloor 4000\sqrt{\frac{33}{53}} \rfloor = \boxed{3156}$ .

Let the centers of the ellipse and the circle be $O(0,0)$ and $Q(x_Q, 0)$ , and the points the circle is tangent to $y = 2x$ and the ellipse be $P$ and $N$ respectively..

As $\tan \angle POQ = 2$ , $OQ = PQ \cdot \csc \angle POQ \implies x_Q = \dfrac {\sqrt 5}2R$ . Then the equation of the circle and its gradient at a point $(x,y)$ are:

$\begin{aligned} \left(x - \frac {\sqrt 5 R}2 \right)^2 + y^2 & = R^2 & ...(1a) \\ 2\left(x - \frac {\sqrt 5 R}2 \right) + 2y \frac {dy}{dx} & = 0 \\ \implies \frac {dy}{dx} & = - \frac {x-\frac {\sqrt 5 R}2}y & ...(1b) \end{aligned}$

Similarly,

$\begin{aligned} \frac {x^2}{49} + \frac {y^2}{16} & = 1 & ...(2a) \\ \frac {2x}{49} + \frac {2y}{16}\cdot \frac {dy}{dx} & = 0 \\ \implies \frac {dy}{dx} & = - \frac {16x}{49y} & ...(2b) \end{aligned}$

At $R$ , we have $(1a) = (2a)$ and $(1a) = (2b)$ .

$\begin{aligned} (1b) = (2b): \quad - \frac {x-\frac {\sqrt 5 R}2}y &= - \frac {16x}{49y} \implies x = \frac {49\sqrt 5 R}{66} \end{aligned}$

$\begin{aligned} 16 \times (2a) -(1a): \quad \sqrt 5 R \blue x - \frac 54 R^2 & = 16-R^2 & \small \blue{\text{Note that }x = \frac {49\sqrt 5 R}{66}} \\ R^2 & = \frac {16 \times 33}{53} \\ \implies R & = 4\sqrt{\frac {33}{53}} \approx 3.15630545907 \\ \lfloor 10^4 R \rfloor & = \boxed{3156} \end{aligned}$