Circumcenter Geometry

By cosine rule , we have:

$\begin{cases} \triangle ABC: & 15^2 = 14^2 + 13^2 - 2(14)(13)\cos A & ...(1) \\ \triangle AXC: & |CX|^2 = 28^2 + 13^2 - 2(28)(13)\cos A & ...(2) \end{cases}$

$\begin{aligned} (2)-2(1): \quad |CX|^2 - 2(15^2) & = 2(14^2) - 13^2 \\ \implies |CX| & = \sqrt {673} \end{aligned}$

Since the chord $AB$ and chord $CD$ meet externally at $X$ , we have:

$\begin{aligned} |CX|\times |DX| & = |AX| \times |BX| \\ \sqrt {673}(\sqrt {673}-|CD|) & = 28(14) \\ \implies |CD| & = \frac {281}{\sqrt {673}} \end{aligned}$

Therefore $a+b=281+673 = \boxed{954}$ .

$\overline {CB}$ is the median of $\triangle {ACX}$ . So $(\overline {AC})^2+(\overline {CX})^2=2(\overline {BX})^2+2(\overline {CB})^2$ , or $\overline {CX}=\sqrt {673}$ . Now, the chords $\overline {AB}$ and $\overline {CD}$ meet at $X$ . So $|\overline {CX}|\times |\overline {DX}|=|\overline {AX}|\times |\overline {BX}|=28\times 14=392$ . Or $\sqrt {673}\times (\sqrt {673}-|\overline {CD}|)=392$ , or $\sqrt {673}\times |\overline {CD}|=673-392=281$ . Therefore $|\overline {CD}|=\dfrac{281}{\sqrt {673}}$ . Hence $a=281, b=673$ and $a+b=281+673=\boxed {954}$

Let $E$ be the foot of the altitude from $C$ to $AB$ . By Heron's, we find that $[ABC]=84$ , or $CE=12$ . We can see that $\triangle ACE$ is a $5-12-13$ triangle, so $AE=5$ . Then $XC=\sqrt{12^2+23^2}=\sqrt{673}$ . Let $XD=p$ and $DC=q$ . Then $p+q=\sqrt{673}$ and, by PoP, $p(p+q)=14\cdot28=392\rightarrow p\sqrt{673}=392$ , so $p=\frac{392}{\sqrt{673}}$ , and $q=\sqrt{673}-p=\frac{281}{\sqrt{673}}$ , so our answer is $281+673=\boxed{954}$ .