Step 1: Determine the Coordinates of the Focus S of the Hyperbola

The first crucial step is to identify the properties of the given hyperbola and determine the coordinates of its focus, S.

Key Concept: Standard Form and Foci of a Hyperbola The standard form of a hyperbola centered at the origin with its transverse axis along the $x$-axis is given by: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ For such a hyperbola, the foci are located at $(\pm ae, 0)$, where $a$ is the semi-transverse axis length, $b$ is the semi-conjugate axis length, and $e$ is the eccentricity. The eccentricity $e$ is related to $a$ and $b$ by the formula: $e^2 = 1 + \frac{b^2}{a^2}$

Working: We are given the equation of the hyperbola: $\frac{x^2}{3} - \frac{y^2}{5} = 1$ Explanation: We compare this equation with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ to identify the values of $a^2$ and $b^2$. From the given equation, we have: $a^2 = 3 \implies a = \sqrt{3} \quad (\text{since } a > 0)$ $b^2 = 5$ Now, we calculate the eccentricity $e$. Explanation: The eccentricity is a measure of how "stretched" the hyperbola is, and it's essential for finding the foci. We use the formula $e^2 = 1 + \frac{b^2}{a^2}$: $e^2 = 1 + \frac{5}{3}$ $e^2 = \frac{3+5}{3} = \frac{8}{3}$ Taking the square root to find $e$: $e = \sqrt{\frac{8}{3}} \quad (\text{since } e > 1 \text{ for a hyperbola})$ The foci of the hyperbola are at $(\pm ae, 0)$. Explanation: The problem explicitly states that S is the focus on the positive $x$-axis. Therefore, we choose the positive value for $ae$: $S = (ae, 0) = \left( \sqrt{3} \cdot \sqrt{\frac{8}{3}}, 0 \right)$ To simplify the $x$-coordinate: $S = \left( \sqrt{3 \cdot \frac{8}{3}}, 0 \right) = (\sqrt{8}, 0)$ Further simplifying $\sqrt{8}$: $S = (2\sqrt{2}, 0)$ Thus, the coordinates of the focus $S$ are $(2\sqrt{2}, 0)$.

Tip for Success: Always be mindful of the difference in the eccentricity formula for hyperbolas ($e^2 = 1 + \frac{b^2}{a^2}$) and ellipses ($e^2 = 1 - \frac{b^2}{a^2}$ if $a^2 > b^2$, or $e^2 = 1 - \frac{a^2}{b^2}$ if $b^2 > a^2$). A common mistake is to interchange these.

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Step 2: Determine the Coordinates of Point B

Next, we utilize the information about the circle and its diameter to find the coordinates of point B.

Key Concept: Midpoint of a Diameter If a line segment $SAB$ is a diameter of a circle and $A$ is the center of that circle, then $A$ must be the midpoint of the segment $SB$. The midpoint formula for two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is $M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$. If we know the midpoint $M(x_M, y_M)$ and one endpoint $P_1(x_1, y_1)$, the other endpoint $P_2(x_2, y_2)$ can be found using: $x_2 = 2x_M - x_1$ $y_2 = 2y_M - y_1$

Working: We are given:

• The center of the circle $C$ is $A(\sqrt{6}, \sqrt{5})$.
• $SAB$ is a diameter of $C$. This means $A$ is the midpoint of $SB$.
• From Step 1, we found the coordinates of $S(2\sqrt{2}, 0)$. Explanation: Since $A$ is the center of the circle and $SAB$ is a diameter, $A$ serves as the midpoint for the segment connecting $S$ and $B$. Let the coordinates of point $B$ be $(x_B, y_B)$. We can use the rearranged midpoint formula to solve for $x_B$ and $y_B$. Using the midpoint formula where $A$ is the midpoint of $SB$: $x_A = \frac{x_S + x_B}{2} \implies x_B = 2x_A - x_S$ $y_A = \frac{y_S + y_B}{2} \implies y_B = 2y_A - y_S$ Substitute the known coordinates of $S(2\sqrt{2}, 0)$ and $A(\sqrt{6}, \sqrt{5})$: For the $x$-coordinate of $B$: $x_B = 2(\sqrt{6}) - 2\sqrt{2}$ $x_B = 2\sqrt{6} - 2\sqrt{2}$ For the $y$-coordinate of $B$: $y_B = 2(\sqrt{5}) - 0$ $y_B = 2\sqrt{5}$ So, the coordinates of point $B$ are $(2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})$.

Common Mistake: When working with expressions involving square roots, ensure you perform arithmetic operations correctly. For example, $2\sqrt{6} - 2\sqrt{2}$ cannot be simplified further as the radicands (6 and 2) are different.

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Step 3: Calculate the Area of Triangle OSB

Now that we have the coordinates of all three vertices, we can calculate the area of triangle OSB.

Key Concept: Area of a Triangle with Vertices The area of a triangle with vertices $O(0,0)$, $S(x_S, y_S)$, and $B(x_B, y_B)$ can be calculated using the determinant formula: $\text{Area}(\Delta OSB) = \frac{1}{2} \left| x_S y_B - x_B y_S \right|$ Alternatively, if one side of the triangle lies on an axis, we can use the simpler base-height formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Working: The vertices of the triangle OSB are:

• $O = (0,0)$ (the origin)
• $S = (2\sqrt{2}, 0)$
• $B = (2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})$ Explanation: Notice that point $S(2\sqrt{2}, 0)$ lies on the $x$-axis. This means the side $OS$ of the triangle lies along the $x$-axis. This significantly simplifies the area calculation, as we can consider $OS$ as the base and the perpendicular distance from $B$ to the $x$-axis as the height. The length of the base $OS$ is the distance from $O(0,0)$ to $S(2\sqrt{2}, 0)$, which is simply $|2\sqrt{2} - 0| = 2\sqrt{2}$. The height of the triangle with respect to the base $OS$ is the absolute value of the $y$-coordinate of the third vertex, $B$. So, the height is $|2\sqrt{5}| = 2\sqrt{5}$. Using the base-height formula: $\text{Area}(\Delta OSB) = \frac{1}{2} \times \text{base} \times \text{height}$ $\text{Area}(\Delta OSB) = \frac{1}{2} \times (2\sqrt{2}) \times (2\sqrt{5})$ $\text{Area}(\Delta OSB) = \frac{1}{2} \times (2 \times 2) \times (\sqrt{2} \times \sqrt{5})$ $\text{Area}(\Delta OSB) = \frac{1}{2} \times 4\sqrt{10}$ $\text{Area}(\Delta OSB) = 2\sqrt{10}$

Alternative using the determinant formula: $\text{Area}(\Delta OSB) = \frac{1}{2} \left| x_S y_B - x_B y_S \right|$ $\text{Area}(\Delta OSB) = \frac{1}{2} \left| (2\sqrt{2})(2\sqrt{5}) - (2\sqrt{6} - 2\sqrt{2})(0) \right|$ $\text{Area}(\Delta OSB) = \frac{1}{2} \left| 4\sqrt{10} - 0 \right|$ $\text{Area}(\Delta OSB) = \frac{1}{2} \cdot 4\sqrt{10} = 2\sqrt{10}$ Both methods confirm the area is $2\sqrt{10}$.

Tip for Efficiency: When one vertex of a triangle is the origin $(0,0)$, or when one side lies along an axis, the base-height formula or the simplified determinant formula $\frac{1}{2} |x_1 y_2 - x_2 y_1|$ is generally faster and less error-prone than the general $3 \times 3$ determinant formula for triangle area.

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Step 4: Calculate the Square of the Area of Triangle OSB

The final step is to fulfill the specific requirement of the question: to find the square of the calculated area.

Key Concept: Squaring Expressions with Square Roots To find the square of an expression of the form $(k\sqrt{m})$, we apply the property $(xy)^n = x^n y^n$: $(k\sqrt{m})^2 = k^2 \cdot (\sqrt{m})^2 = k^2 m$

Working: From Step 3, we found that the Area of $\Delta OSB = 2\sqrt{10}$. Explanation: The problem asks for the square of this area. We apply the squaring operation to our result. $(\text{Area}(\Delta OSB))^2 = (2\sqrt{10})^2$ $(\text{Area}(\Delta OSB))^2 = 2^2 \cdot (\sqrt{10})^2$ $(\text{Area}(\Delta OSB))^2 = 4 \cdot 10$ $(\text{Area}(\Delta OSB))^2 = 40$

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Summary and Key Takeaways

This problem serves as an excellent test of multiple fundamental concepts in coordinate geometry and conic sections. To solve it efficiently and accurately, one must:

1. Master Hyperbola Properties: Be proficient in identifying $a^2$, $b^2$, calculating eccentricity ($e$), and finding the foci coordinates for a hyperbola. Remember the specific eccentricity formula for hyperbolas.
2. Understand Circle Geometry and Midpoint Formula: Recognize that the center of a circle is the midpoint of any of its diameters. Be able to apply the midpoint formula effectively, including rearranging it to find an endpoint when the midpoint and the other endpoint are known.
3. Choose Efficient Area Calculation Methods: For a triangle with one vertex at the origin or a side along an axis, the base-height method or the simplified determinant formula is much quicker than the general $3 \times 3$ determinant.
4. Perform Algebraic Operations Carefully: Pay close attention to calculations involving square roots to avoid errors.

By systematically applying these concepts and being meticulous with calculations, this problem can be solved effectively.

The final answer is $\boxed{40}$.