Problem 1459

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:
$$\boxed{13}$$