Sun Embracing the Moon

The radius of the larger circle is $25$

Since $AC\perp PQ$ and $AP=AQ$ ,

we have $PC=CQ=7$

Similarly, $QD=DR=20$

$\cos \angle PQA=\cfrac{7}{25}$

$\Rightarrow \tan \angle PQA=\cfrac{24}{7}$

$\cos \angle RQA=\cfrac{20}{25}=\cfrac{4}{5}$

$\Rightarrow \tan \angle RQA=\cfrac{3}{4}$

Let $\angle PQA=\alpha, \angle RQA=\beta$

$\Rightarrow \tan \alpha=\cfrac{24}{7}, \tan \beta=\cfrac{3}{4}$

$\tan(\alpha+\beta)=\cfrac{\cfrac{24}{7}+\cfrac{3}{4}}{1-\cfrac{24}{7}\cdot\cfrac{3}{4}}=-\cfrac{117}{44}$

Let $\alpha+\beta=2\gamma$

$\Rightarrow \tan(\alpha+\beta)=\tan2\gamma$

$\cfrac{2\tan\gamma}{1-(\tan\gamma)^2}=-\cfrac{117}{44}$

$-88\tan\gamma=117-117(\tan\gamma)^2$

$117(\tan\gamma)^2-88\tan\gamma-117=0$

$(13\tan\gamma+9)(9\tan\gamma-13)=0$

$\tan\gamma=-\cfrac{9}{13}$ or $\tan\gamma=\cfrac{13}{9}$

$\because 0<\gamma<\cfrac{\pi}{2}$

$\therefore \tan\gamma>0$

$\therefore \tan\gamma=-\cfrac{9}{13}$ is unsuitable

$\therefore \tan\gamma=\cfrac{13}{9}$

$\angle PQF=\angle FQR=\gamma$

$\therefore \tan \angle PQF=\cfrac{13}{9}$

$\sin\angle PQF=\cfrac{13}{\sqrt{13^2+9^2}}=\cfrac{13}{\sqrt{250}}=\cfrac{13}{5\sqrt{10}}$

$\sin\angle PQF=\cfrac{r}{QF}$

$\therefore \cfrac{r}{QF}=\cfrac{13}{5\sqrt{10}}$

$QF=\cfrac{5\sqrt{10}r}{13}$

$QA=25$

$\because$ the two circles are internally tangent

$\therefore$ the two centers $A,F$ and the tangent point $E$ are collinear.

$\therefore FA=25-r$

$\angle FQA=\alpha-\gamma$

$\tan \angle FQA=\tan(\alpha-\gamma)=\cfrac{\tan\alpha-\tan\gamma}{1+\tan\alpha\tan\gamma}=\cfrac{\cfrac{24}{7}-\cfrac{13}{9}}{1+\cfrac{24}{7}\cdot\cfrac{13}{9}}=\cfrac{125}{375}=\cfrac{1}{3}$

$\cos \angle FQA=\cfrac{3}{\sqrt{1^2+3^2}}=\cfrac{3}{\sqrt{10}}$

By Cosine Rule ,

$(25-r)^2=(\cfrac{5\sqrt{10}r}{13})^2+25^2-2\cdot\cfrac{5\sqrt{10}r}{13}\cdot25\cdot\cfrac{3}{\sqrt{10}}$

$(25-r)^2=\cfrac{250r^2}{13^2}+25^2-\cfrac{750r}{13}$

$169(25-r)^2=250r^2+25^2\cdot169-750r\cdot13$

$169r^2-169\cdot50r+169\cdot25^2=250r^2+25^2\cdot169-750r\cdot13$

$169r^2-169\cdot50r=250r^2-750r\cdot13$

$81r^2+(169\cdot50-750\cdot13)r=0$

$81r^2+13\cdot50(13-15)=0$

$81r^2-2\cdot13\cdot50=0$

$81r^2-1300r=0$

$r(81r-1300)=0$

$r=0$ or $r=\cfrac{1300}{81}$

$\because r>0$

$\therefore r=0$ is unsuitable

$r=\cfrac{1300}{81}$

$81r=\boxed{1300}$

Let the center of the larger circle be $S$ and draw three radii $SV$ (the perpendicular bisector of chord $PQ$ with intersection $T$ ), $SW$ (the perpendicular bisector of chord $QR$ with intersection $U$ ), and $SQ$ .

Then as radii of the larger circle, $SV = SW = SQ = \frac{1}{2}50 = 25$ . Also, $QT = \frac{1}{2}14 = 7$ and $QU = \frac{1}{2}40 = 20$ . By Pythagorean's Theorem, $SU = \sqrt{25^2 - 20^2} = 15$ and $ST = \sqrt{25^2 - 7^2} = 24$ , and $\angle QSU = \tan^{-1}(\frac{20}{15}) = \tan^{-1}(\frac{4}{3})$ and $\angle QST = \tan^{-1}(\frac{7}{24})$ . Using the tangent angle addition formula, $\tan \angle TSU$ $=$ $\frac{\frac{4}{3} + \frac{7}{24}}{1 - \frac{4}{3} \frac{7}{24}}$ $=$ $\frac{117}{44}$ , and so $\angle TSU = \tan ^{-1} (\frac{117}{44})$ .

Now let the center of the smaller circle be $M$ and draw two radii $MN$ (where $N$ is the point of tangency of the small circle and chord $QR$ ) and $MO$ (where $O$ is the point of tangency of the small circle and chord $PQ$ ).

Since $MN$ and $SU$ are perpendicular to the same line $PQ$ , they are parallel, and since $MO$ and $ST$ are perpendicular to the same line $QR$ , they are parallel, which means $\angle OMN = \angle TSU = \tan ^{-1} (\frac{117}{44})$ .

Finally, let $K$ be the point of tangency between the two circles and draw $SK$ (through $M$ ). Also draw $MQ$ and $SL$ where $SL$ is perpendicular to $MN$ at $L$ .

Since $MO$ and $MN$ are radii of circle $M$ , $MO \cong MN$ , and since $\triangle QMO$ and $\triangle QMN$ are perpendicular triangles with a shared side $QM$ , $\triangle QMO \cong \triangle QMN$ by hypotenuse-leg congruency. Therefore, $\angle QMN = \angle QMO$ $=$ $\frac{1}{2} \angle OMN$ $=$ $\frac{1}{2} \tan ^{-1} (\frac{117}{44})$ . By tangent half angle formula, $\tan \angle QMN$ $=$ $\tan (\frac{1}{2} \tan ^{-1} (\frac{117}{44}))$ $=$ $\frac{\sin (\tan ^{-1} (\frac{117}{44}))}{1 + \cos (\tan ^{-1} (\frac{117}{44}))}$ $=$ $\frac{\frac{117}{125}}{1 + \frac{44}{125}} = \frac{9}{13}$ .

Let $r$ be the radius of the smaller circle. Then $MN = r$ and $QN = r \tan \angle QMN = \frac{9}{13}r$ , and the sides of $\triangle MLS$ are $SM = SK - MK = 25 - r$ , $LM = MN - SU = r - 15$ , and $LS = QU - QN = 20 - \frac{9}{13}r$ . Since $\triangle MLS$ is a right triangle, by Pythagorean Theorem $LM^2 + LS^2 = SM^2$ , so $(r - 15)^2 + (20 - \frac{9}{13}r)^2 = (25 - r)^2$ , and solving for $r > 0$ gives $r = \frac{1300}{81}$ . Therefore, $81r = 81 \cdot \frac{1300}{81} = \boxed{1300}$ .