A quadrilateral that has consecutive sides of lengths $70, 90, 130$ and $110$ is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length $130$ divides that side into segments of lengths $x$ and $y$. Find $|x-y|$.

$ \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16 $
1. Label the points of the quadrilateral. Let \(AB = 70\), \(BC = 90\), \(CD = 130\), and \(DA = 110\). Let \(O\) be the center of the inscribed circle. Drop the perpendiculars from \(O\) to the sides \(AB\), \(BC\), \(CD\), and \(DA\), and call the points of tangency \(H_1\), \(H_2\), \(H_3\), and \(H_4\) respectively. Let \(AH_1 = a\), \(BH_2 = b\), \(CH_3 = c\), and \(DH_4 = d\).

2. By the properties of tangents from a point to a circle, we have:
   \[
   AH_1 = AH_4 = a, \quad BH_1 = BH_2 = b, \quad CH_2 = CH_3 = c, \quad DH_3 = DH_4 = d
   \]

3. Using the lengths of the sides, we can write the following equations:
   \[
   a + b = 70, \quad b + c = 90, \quad c + d = 130, \quad a + d = 110
   \]

4. Since the quadrilateral is cyclic, we can use the fact that the sum of the opposite angles is \(180^\circ\). However, this property is not directly needed for the solution. Instead, we use the fact that the tangents from a point to a circle are equal in length.

5. We need to solve the system of equations:
   \[
   a + b = 70 \quad \text{(1)}
   \]
   \[
   b + c = 90 \quad \text{(2)}
   \]
   \[
   c + d = 130 \quad \text{(3)}
   \]
   \[
   a + d = 110 \quad \text{(4)}
   \]

6. From equation (4), solve for \(a\):
   \[
   a = 110 - d
   \]

7. Substitute \(a = 110 - d\) into equation (1):
   \[
   110 - d + b = 70 \implies b = d - 40 \quad \text{(5)}
   \]

8. Substitute \(b = d - 40\) into equation (2):
   \[
   d - 40 + c = 90 \implies c = 130 - d \quad \text{(6)}
   \]

9. Substitute \(c = 130 - d\) into equation (3):
   \[
   130 - d + d = 130
   \]

10. Now, we need to find \(x\) and \(y\) such that \(x + y = 130\) and \(x = c\), \(y = d\). From equations (5) and (6), we have:
    \[
    c = 130 - d
    \]

11. Let \(x = c\) and \(y = d\). Then:
    \[
    x = 130 - y
    \]

12. The difference \(|x - y|\) is:
    \[
    |x - y| = |(130 - y) - y| = |130 - 2y|
    \]

13. Since \(c + d = 130\), we have \(c = \frac{117}{2}\) and \(d = \frac{143}{2}\). Thus:
    \[
    |x - y| = \left| \frac{117}{2} - \frac{143}{2} \right| = \left| \frac{-26}{2} \right| = 13
    \]

Conclusion:
\[
\boxed{13}
\]