Key Concepts and Formulas

• Equation of a Circle: The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
• Distance from a Point to a Line: The perpendicular distance from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is given by $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
• Geometric Property: The line joining the center of a circle to the point of tangency is perpendicular to the tangent at that point.

Step-by-Step Solution

Step 1: Identify the given information and derive the radius.

We are given the circle $(x-\alpha)^2 + (y-\beta)^2 = 50$, where $\alpha, \beta > 0$. The center of the circle is $(\alpha, \beta)$ and the radius is $r = \sqrt{50} = 5\sqrt{2}$.

Step 2: Use the distance of the point of tangency from the origin.

Let the point of tangency be $P(x_1, y_1)$. We are given that the distance of $P$ from the origin is $4\sqrt{2}$. Therefore, $x_1^2 + y_1^2 = (4\sqrt{2})^2 = 32$.

Step 3: Use the fact that P lies on the line y + x = 0.

Since $P(x_1, y_1)$ lies on the line $y + x = 0$, we have $y_1 = -x_1$. Substituting this into the equation from Step 2, we get $x_1^2 + (-x_1)^2 = 32$, which simplifies to $2x_1^2 = 32$. Thus, $x_1^2 = 16$, so $x_1 = \pm 4$.

Step 4: Find the coordinates of the point of tangency P.

If $x_1 = 4$, then $y_1 = -4$, so $P = (4, -4)$. If $x_1 = -4$, then $y_1 = 4$, so $P = (-4, 4)$.

Step 5: Consider the slope of the line joining the center to the point of tangency.

The slope of the tangent line $y + x = 0$ is $-1$. The line joining the center $(\alpha, \beta)$ to the point of tangency is perpendicular to the tangent. Therefore, the slope of the line joining the center to the point of tangency is $1$.

Step 6: Use the slope to relate alpha and beta.

Case 1: $P = (4, -4)$. The slope of the line joining $(\alpha, \beta)$ to $(4, -4)$ is $\frac{\beta - (-4)}{\alpha - 4} = \frac{\beta + 4}{\alpha - 4} = 1$. This gives $\beta + 4 = \alpha - 4$, so $\alpha - \beta = 8$.

Case 2: $P = (-4, 4)$. The slope of the line joining $(\alpha, \beta)$ to $(-4, 4)$ is $\frac{\beta - 4}{\alpha - (-4)} = \frac{\beta - 4}{\alpha + 4} = 1$. This gives $\beta - 4 = \alpha + 4$, so $\beta - \alpha = 8$. Since $\alpha, \beta > 0$, both cases are valid.

Step 7: Use the distance from the center to the tangent line.

The distance from the center $(\alpha, \beta)$ to the line $x + y = 0$ is equal to the radius $5\sqrt{2}$. Therefore, $\frac{|\alpha + \beta|}{\sqrt{1^2 + 1^2}} = 5\sqrt{2}$, which simplifies to $\frac{|\alpha + \beta|}{\sqrt{2}} = 5\sqrt{2}$. This means $|\alpha + \beta| = 10$, so $\alpha + \beta = 10$ (since $\alpha, \beta > 0$).

Step 8: Solve for alpha and beta.

Case 1: $\alpha - \beta = 8$ and $\alpha + \beta = 10$. Adding these equations gives $2\alpha = 18$, so $\alpha = 9$. Then $\beta = 10 - 9 = 1$.

Case 2: $\beta - \alpha = 8$ and $\alpha + \beta = 10$. Adding these equations gives $2\beta = 18$, so $\beta = 9$. Then $\alpha = 10 - 9 = 1$.

Step 9: Calculate (alpha + beta)^2.

In both cases, $\alpha + \beta = 10$. Therefore, $(\alpha + \beta)^2 = 10^2 = 100$.

Step 10: Re-evaluate the problem.

The value of $r^2$ is 50. So the distance from the center $(\alpha,\beta)$ to the tangent $x+y=0$ must be $5\sqrt{2}$. This implies $|\alpha+\beta|/\sqrt{2} = 5\sqrt{2}$ which means $|\alpha+\beta|=10$. Since $\alpha, \beta >0$, we get $\alpha+\beta=10$. Also, the line joining the center $(\alpha,\beta)$ to the point of contact $(x_1, y_1)$ is perpendicular to the tangent. Since the slope of the tangent $x+y=0$ is $-1$, the slope of the perpendicular is 1. This means $\frac{\beta-y_1}{\alpha-x_1} = 1$. Since $P(x_1,y_1)$ is at a distance $4\sqrt{2}$ from the origin, and since $P$ lies on the line $x+y=0$, we can write $y_1=-x_1$. So $x_1^2 + y_1^2 = 32$ gives $x_1^2 + x_1^2=32$, so $x_1^2=16$ and $x_1=\pm 4$. Thus $P$ is either $(4,-4)$ or $(-4,4)$.

If $P=(4,-4)$, we have $\frac{\beta+4}{\alpha-4}=1$ so $\beta+4=\alpha-4$ or $\alpha-\beta=8$. If $P=(-4,4)$, we have $\frac{\beta-4}{\alpha+4}=1$ so $\beta-4=\alpha+4$ or $\beta-\alpha=8$. We also have $\alpha+\beta=10$. In the first case, we have $\alpha-\beta=8$ and $\alpha+\beta=10$. Adding gives $2\alpha=18$ so $\alpha=9$ and $\beta=1$. In the second case, we have $\beta-\alpha=8$ and $\alpha+\beta=10$. Adding gives $2\beta=18$ so $\beta=9$ and $\alpha=1$. In both cases, $(\alpha,\beta)=(9,1)$ or $(1,9)$. The question asks for $(\alpha+\beta)^2$ which equals $10^2=100$.

Step 11: Identify and correct the error.

The question asks for $(\alpha+\beta)^2$. The distance from the center $(\alpha, \beta)$ to the tangent $x + y = 0$ is $5\sqrt{2}$. Thus $|\alpha + \beta| / \sqrt{2} = 5\sqrt{2}$, which implies $|\alpha + \beta| = 10$. Since $\alpha, \beta > 0$, $\alpha + \beta = 10$. Therefore $(\alpha + \beta)^2 = 100$. However, the correct answer provided is 2. Let us assume $\alpha=\beta$. Then $(\alpha-\alpha)^2 + (\beta-\beta)^2=50$ becomes $2(\alpha-\beta)^2 = 0$. Since $P$ lies on $x+y=0$, $P=(-4,4)$ or $(4,-4)$. Then $\alpha=1$, $\beta=9$ or $\alpha=9$, $\beta=1$. So $(\alpha+\beta)^2 = 100$. The distance from $(\alpha,\beta)$ to the line $x+y=0$ is $\frac{|\alpha+\beta|}{\sqrt{2}}$. This equals $5\sqrt{2}$. So $|\alpha+\beta|=10$. Also, $\alpha-\beta=8$ or $\beta-\alpha=8$. So $\alpha=9$, $\beta=1$ or $\beta=9$, $\alpha=1$. Then $\alpha+\beta=10$. $(\alpha+\beta)^2=100$. There is an error in the question.

The correct answer is 100. However, according to the "Correct Answer" given, the answer should be 2. There must be an error in the problem statement or the provided solution.

Since the given "Correct Answer" is 2, and we are asked to find $(\alpha+\beta)^2$, it implies that $\alpha+\beta=\sqrt{2}$.

However, $\frac{|\alpha+\beta|}{\sqrt{2}}=5\sqrt{2}$, hence $|\alpha+\beta|=10$. Therefore, $(\alpha+\beta)^2=100$.

Common Mistakes & Tips

• Carefully check the signs when calculating the slope and using the distance formula.
• Remember that the distance is always positive, so use the absolute value when applying the distance from a point to a line formula.
• Be mindful of the given conditions, such as $\alpha, \beta > 0$, to eliminate extraneous solutions.

Summary

We determined the center and radius of the circle. Using the distance of the point of tangency from the origin and the fact that it lies on the line $y+x=0$, we found the coordinates of the point of tangency. We used the perpendicularity condition between the radius and the tangent to establish a relationship between $\alpha$ and $\beta$. Finally, we used the distance from the center to the tangent line to obtain another relationship between $\alpha$ and $\beta$. Solving these equations, we found $(\alpha + \beta)^2 = 100$.

Final Answer

The final answer is \boxed{100}.