1. **State the problem:** We have two congruent circles with centers $P$ and $Q$. Points $R$, $P$, $Q$, and $S$ lie on a horizontal line such that $RP = QS = 1$. The circles intersect at points $X$ and $Y$, and the length of segment $XY$ is 8. We need to find the distance $PQ$ between the centers. 2. **Analyze the setup:** Since the circles are congruent, their radii are equal. Let the radius be $r$. Points $R$ and $S$ lie on the circles such that $RP = QS = 1$, so $R$ is 1 unit left of $P$ and $S$ is 1 unit right of $Q$. 3. **Identify the radius:** Because $R$ lies on the circle centered at $P$, the radius $r = PR = 1 + RP = 1 + 1 = 2$ is incorrect since $RP=1$ is the distance from $R$ to $P$. Actually, $RP=1$ means $R$ is 1 unit from $P$, so $r = PR = 1$. Similarly, $QS=1$ means $S$ is 1 unit from $Q$, so $r = QS = 1$. Therefore, the radius of each circle is $r=1$. 4. **Set coordinate system:** Place $P$ at the origin $(0,0)$ on the horizontal axis. Then $R$ is at $(-1,0)$ since $RP=1$. Since $PQ$ is the distance we want to find, let $Q$ be at $(d,0)$ where $d = PQ$. 5. **Coordinates of points $X$ and $Y$:** Points $X$ and $Y$ lie on both circles, so they satisfy both circle equations: Circle $P$: $x^2 + y^2 = r^2 = 1$ Circle $Q$: $(x - d)^2 + y^2 = 1$ 6. **Find intersection points:** Subtract the two equations: $$x^2 + y^2 = 1$$ $$ (x - d)^2 + y^2 = 1$$ Subtracting gives: $$x^2 + y^2 - ((x - d)^2 + y^2) = 1 - 1$$ $$x^2 - (x - d)^2 = 0$$ Expand: $$x^2 - (x^2 - 2dx + d^2) = 0$$ $$x^2 - x^2 + 2dx - d^2 = 0$$ $$2dx = d^2$$ $$x = \frac{d}{2}$$ 7. **Find $y$ coordinate:** Substitute $x = \frac{d}{2}$ into the first circle equation: $$\left(\frac{d}{2}\right)^2 + y^2 = 1$$ $$\frac{d^2}{4} + y^2 = 1$$ $$y^2 = 1 - \frac{d^2}{4}$$ 8. **Use length $XY = 8$:** Points $X$ and $Y$ have the same $x$ coordinate $\frac{d}{2}$ and $y$ coordinates $y$ and $-y$. So, $$XY = |y - (-y)| = 2|y| = 8$$ Therefore, $$|y| = 4$$ 9. **Solve for $d$:** From step 7, $$y^2 = 1 - \frac{d^2}{4} = 16$$ So, $$1 - \frac{d^2}{4} = 16$$ $$- \frac{d^2}{4} = 15$$ $$d^2 = -60$$ This is impossible, so our assumption about radius $r=1$ is incorrect. 10. **Re-examine radius:** Since $RP = 1$ and $R$ lies on the circle centered at $P$, $PR$ is radius $r$, so $r = RP = 1$. But this contradicts the previous step. The problem states $RP = QS = 1$, but $R$ and $S$ are points on the horizontal line through $P$ and $Q$, not necessarily on the circle. 11. **Correct interpretation:** $R$ and $S$ lie on the line through $P$ and $Q$, with $RP = QS = 1$, but $R$ and $S$ are points on the circles, so the radius $r$ is the distance from $P$ to $R$ and from $Q$ to $S$. Therefore, radius $r = PR = 1$ and $r = QS = 1$. 12. **Recalculate with radius $r=1$:** From step 7, $$y^2 = r^2 - \frac{d^2}{4} = 1 - \frac{d^2}{4}$$ From step 8, $$2|y| = 8 \Rightarrow |y| = 4$$ So, $$y^2 = 16$$ Set equal: $$16 = 1 - \frac{d^2}{4}$$ $$- \frac{d^2}{4} = 15$$ $$d^2 = -60$$ Again impossible. 13. **Conclusion:** The radius cannot be 1. Since $RP = QS = 1$ and $R$ and $S$ lie on the circles, the radius $r$ must be $RP + PR$ or $QS + QS$? No, $RP$ and $QS$ are segments from $R$ to $P$ and $Q$ to $S$ respectively, but $R$ and $S$ are points on the circles, so $PR = r$ and $QS = r$. Given $RP = QS = 1$, and $R$ and $S$ lie on the circles, the distance from $P$ to $R$ is $r$, but $RP=1$ means $R$ is 1 unit from $P$, so $r=1$. But this contradicts the length $XY=8$. 14. **Alternative approach:** Let the distance $PQ = d$, radius $r$, and $RP = QS = 1$. Since $R$ lies on the circle centered at $P$, $PR = r$. But $RP = 1$ means $R$ is 1 unit from $P$, so $r = 1$. Similarly, $QS = 1$ means $r = 1$. But $XY = 8$ is the length of the chord formed by the intersection of the two circles. 15. **Formula for chord length of intersecting circles:** The length of the chord $XY$ is given by $$XY = 2 \sqrt{r^2 - \left(\frac{d}{2}\right)^2}$$ Given $XY = 8$, we have $$8 = 2 \sqrt{r^2 - \frac{d^2}{4}}$$ Divide both sides by 2: $$4 = \sqrt{r^2 - \frac{d^2}{4}}$$ Square both sides: $$16 = r^2 - \frac{d^2}{4}$$ Rearranged: $$r^2 = 16 + \frac{d^2}{4}$$ 16. **Use $RP = QS = 1$ to relate $r$ and $d$:** Since $R$ lies on the circle centered at $P$ and $RP = 1$, and $R$ lies on the line segment $RS$ which passes through $P$ and $Q$, the distance between $P$ and $Q$ is $$d = PQ = RP + RS + SQ = 1 + RS + 1 = RS + 2$$ But $RS$ is the segment between $R$ and $S$, which is the distance between the two points on the circles along the horizontal line. Since $R$ and $S$ lie on the circles, the distance $RS$ is the length of the segment between the two circles along the horizontal line. 17. **Calculate $RS$:** The distance between centers is $d$, and the radius is $r$, so the length of the segment $RS$ is $$RS = d - 2$$ Because $RP = QS = 1$, and $R$ and $S$ lie on the circles, the total distance $RS = d - 2$. 18. **Use Pythagoras to find $r$:** The radius $r$ is the distance from $P$ to $R$, which is $$r^2 = RP^2 + XR^2$$ But $RP = 1$, and $XR$ is the vertical distance from $R$ to $X$ or $Y$. Since $XY = 8$, the vertical distance between $X$ and $Y$ is 8, so half the vertical distance from the center line to $X$ or $Y$ is 4. Therefore, the radius is $$r^2 = 1^2 + 4^2 = 1 + 16 = 17$$ 19. **Find $d$ using $r^2 = 16 + \frac{d^2}{4}$:** From step 15, $$r^2 = 16 + \frac{d^2}{4}$$ Substitute $r^2 = 17$: $$17 = 16 + \frac{d^2}{4}$$ $$\frac{d^2}{4} = 1$$ $$d^2 = 4$$ $$d = 2$$ 20. **Final answer:** The distance between centers $P$ and $Q$ is $$\boxed{2}$$