Coordinates

Let AB = b, and AC = c. Moreover, let $[ABX] = A_1$ and $[ACX] = A_2$, where $[\cdot]$ denote "area of".

By Stewart's theorem, we have 

$(13 + 5)(13^2 + 13(5)) = 5b^2 + 13c^2\\ 5b^2 + 13c^2 = 4212$

By basic geometry, we have $\dfrac{A_1}{A_2} = \dfrac{13}5$. 

Recall that the circumradius of a triangle with sides a, b, c and area A is $\dfrac{abc}{4A}$. Then, since the circumradius of triangles ABX and ACX are the same, we have 

$\dfrac{13^2 b}{4A_1} = \dfrac{5(13)c}{4A_2}\\ \dfrac{A_1}{A_2} = \dfrac{13b}{5c}\\ \dfrac{13}5 = \dfrac{13b}{5c}\\ b = c$.

Hence, 

$18b^2 = 18c^2 = 4212\\ b = c = 3 \sqrt{26}$

We calculate the half perimeter of triangle ABC to be $\dfrac{18 + b + c}2 = 3 \sqrt{26} + 9$. Then, by Heron's formula, the area required is:

$\sqrt{(3 \sqrt{26} + 9)(3\sqrt{26} - 9)(9^2)} = 27\sqrt{17}$.