1. Key Concepts and Formulas

• Area of a triangle given two sides and the included angle: $A = \frac{1}{2}ab\sin C$.
• Area of a triangle given its base and corresponding height: $A = \frac{1}{2} \times \text{base} \times \text{height}$.
• Properties of intersecting circles: The line segment joining the centers of two intersecting circles is the perpendicular bisector of their common chord.

2. Step-by-Step Solution

Step 1: Define the circles and their properties.

We are given two circles:

• $C_1: x^2+y^2=25$ with center $O_1(0,0)$ and radius $R_1 = 5$.
• $C_2: (x-\alpha)^2+y^2=16$ with center $O_2(\alpha,0)$ and radius $R_2 = 4$.

The distance between the centers is $O_1O_2 = \alpha$, since $\alpha \in (5,9)$, which means $\alpha > 0$.

Step 2: Analyze the given information and form the triangle.

Let $P$ be one of the intersection points of $C_1$ and $C_2$. We are given that the angle between the radii at $P$ is $\theta = \sin^{-1}\left(\frac{\sqrt{63}}{8}\right)$. Thus, $\sin \theta = \frac{\sqrt{63}}{8}$. We consider the triangle $\triangle O_1PO_2$. The sides are $O_1P = R_1 = 5$, $O_2P = R_2 = 4$, and $O_1O_2 = \alpha$. The angle $\angle O_1PO_2 = \theta$.

Step 3: Calculate the area of $\triangle O_1PO_2$ using the sine formula.

The area $A$ of $\triangle O_1PO_2$ is given by: $A = \frac{1}{2} \times O_1P \times O_2P \times \sin \theta = \frac{1}{2} \times 5 \times 4 \times \frac{\sqrt{63}}{8} = \frac{20\sqrt{63}}{16} = \frac{5\sqrt{63}}{4}$

Step 4: Relate the area to the common chord.

Let $PQ$ be the common chord of $C_1$ and $C_2$, and let $M$ be the midpoint of $PQ$. Then $O_1O_2 \perp PQ$ and $PM = MQ = \frac{\beta}{2}$, where $\beta$ is the length of the common chord. $PM$ is the height of $\triangle O_1PO_2$ with respect to the base $O_1O_2 = \alpha$.

Step 5: Calculate the area of $\triangle O_1PO_2$ using the base and height formula.

The area $A$ of $\triangle O_1PO_2$ is also given by: $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times O_1O_2 \times PM = \frac{1}{2} \times \alpha \times \frac{\beta}{2} = \frac{\alpha \beta}{4}$

Step 6: Equate the area expressions and solve.

We have two expressions for the area $A$: $\frac{\alpha \beta}{4} = \frac{5\sqrt{63}}{4}$ Multiplying both sides by 4, we get: $\alpha \beta = 5\sqrt{63}$ Squaring both sides: $(\alpha \beta)^2 = (5\sqrt{63})^2 = 25 \times 63 = 25 \times (60 + 3) = 1500 + 75 = 1575$

3. Common Mistakes & Tips

• Visualizing the Geometry: Draw a diagram to help understand the relationships between the circles, their centers, and the common chord.
• Using the correct trigonometric identity: Ensure that you use the correct trigonometric identity to find $\cos \theta$ from $\sin \theta$ if needed. However, this problem can be solved without finding $\cos \theta$.
• Properties of common chord: Remember that the line joining the centers of the circles is perpendicular to the common chord and bisects it.

4. Summary

We found the area of the triangle formed by the centers of the two circles and one of their intersection points in two ways: using the sine of the angle between the radii and using the length of the common chord as the height. Equating these two expressions allowed us to find the value of $(\alpha \beta)^2$, which is 1575.

5. Final Answer

The final answer is \boxed{1575}.