Another Circular Circus Act!

Let the centers of the middle circle and the right circle be $O$ and $Q$ respectively, and the tangent point of the two circles with the ellipse be $P(u,v)$ . Note that $OP=PQ = r$ . Since the $x$ -coordinate of $P$ is $u$ , that of $Q$ is $2u$ .

The line (red) joining $O$ , $P$ , and $Q$ is given by $\dfrac {y-v}{x-u} = m$ , where $m$ is the gradient of the line which is perpendicular to the tangent of the ellipse at point $P$ . The gradient of the tangent to the ellipse is given by

$\dfrac {2x}{25} + \dfrac {2y}{121} \cdot \dfrac {dy}{dx} = 0 \implies \dfrac {dy}{dx} = - \dfrac {121 x}{25 y}$

The gradient of the tangent at point $P (u,v)$ is $- \frac {121 u}{25 v}$ and therefore $m = \frac {25 v}{121 u}$ and the equation of the red line is given by $y = \dfrac {25 v}{121 u}x + \dfrac {96}{121} v$ . Then $O = \left(0, \frac {96}{121} v \right)$ and $Q = \left(2u, \frac {146}{121} v \right)$ . We also note that:

$\begin{cases} u^2 + \left(v-\dfrac {96}{121}v \right)^2 = u^2 + \left(\dfrac {25}{121} \right)^2v^2 = r^2 \\ v + \dfrac {25}{121} v = \dfrac {146}{121} v = r \end{cases}$

Therefore,

$\begin{aligned} \blue{u^2} + \left(\frac {25}{121} \right)^2v^2 & = \left(\frac {146}{121} \right)^2v^2 & \small \blue{\text{Since }\frac {u^2}{25} + \frac {v^2}{121} = 1} \\ \blue{25 - \frac {25}{121} v^2} + \left(\frac {25}{121} \right)^2v^2 & = \left(\frac {146}{121} \right)^2v^2 \\ \frac {23716}{121^2} v^2 & = 25 \\ 23716 v^2 & = 366025 \\ v & = \frac {55}{14} \\ \implies r & = \frac {146}{121} v = \frac {365}{77} \end{aligned}$

Therefore $a+b = \boxed{442}$ .

Let the middle circle have a center of $(0, b$ ) so that its equation is $x^2 + (y - b)^2 = r^2$ .

Substituting $x^2 = r^2 - (y - b)^2$ into $\cfrac{x^2}{25} + \cfrac{y^2}{121} = 1$ and solving for $y$ gives $y = \cfrac{121}{96}b \pm \cfrac{11}{96}\sqrt{25b^2 + 96r^2 - 2400}$ .

For the circle to be tangent to the ellipse, $y$ will only have one value, so $\cfrac{11}{96}\sqrt{25b^2 + 96r^2 - 2400} = 0$ , which solves to $b = \cfrac{4}{5}\sqrt{150 - 6r^2}$ , and $y = \cfrac{121}{96}b$ .

Since the center of the outer circle has a $y$ -coordinate of $r$ , and since the tangent point between the ellipse and circle is the midpoint of the center of the middle circle and one of the outer circles, $\cfrac{1}{2}(b + r) = \cfrac{121}{96}b$ , or $r = \cfrac{73}{48}b$ .

Substituting $b = \cfrac{4}{5}\sqrt{150 - 6r^2}$ into $r = \cfrac{73}{48}b$ gives $r = \cfrac{73}{48} \cdot \cfrac{4}{5}\sqrt{150 - 6r^2}$ , which solves to $r = \cfrac{365}{77}$ .

Therefore, $a = 365$ , $b = 77$ , and $a + b = \boxed{442}$ .

If we let the common tangency point in the first quadrant be $(5\sin\theta,11\cos\theta)$ ,

the slope of the normal at the point will be 1) $\frac{-1}{\frac{dy}{dx}}=\frac{25}{121}\frac{11\cos\theta}{5\sin\theta}=\frac{5}{11\tan\theta}=\tan\phi$ ,

Now, moving a distance $r$ to the right along the normal will make the ordinate $=r$

2) $11\cos\theta + r\sin\phi=r$

And, moving $r$ to the left we will end up on the y-axis, since the center of the middle circle has to lie there by symmetry.

3) $5\sin\theta-r\cos\phi=0$

Using the last 2 equations together and the first relation,

$\tan\phi = \frac{r-11\cos\theta}{5\sin\theta}=\frac{5}{11\tan\theta}$ ,

This gives $r=\frac{146\cos\theta}{11}$

Substituting this in 2 gives,

$\cos\theta=\frac{5}{14}$

Therefore, $r=\frac{365}{77}$