Key Concepts and Formulas

• Tangents from an external point to a circle are perpendicular to the radius at the point of tangency.
• The circumcircle of the triangle formed by the origin and the points of tangency has a diameter that extends from the origin to the center of the original circle.
• The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.

Step-by-Step Solution

Step 1: Find the center of the given circle. The given circle equation is $x^2 + y^2 - 6x + 4y + 8 = 0$. We need to rewrite it in the standard form $(x-h)^2 + (y-k)^2 = r^2$ to find the center $(h, k)$. However, it's easier to recognize that for a general equation of the form $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is $(-g, -f)$. Here, $2g = -6$, so $g = -3$. Also, $2f = 4$, so $f = 2$. Therefore, the center of the circle is $C = (-(-3), -2) = (3, -2)$.

Step 2: Determine the equation of the circumcircle of triangle OPQ. Since OP and OQ are tangents to the given circle from the origin O, and P and Q are the points of tangency, the circumcircle of triangle OPQ passes through the origin O(0, 0) and has OC as its diameter. The coordinates of O are (0, 0), and the coordinates of C are (3, -2). The equation of a circle with endpoints of a diameter at $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting the coordinates of O and C: $(x - 0)(x - 3) + (y - 0)(y - (-2)) = 0$ $x(x - 3) + y(y + 2) = 0$ $x^2 - 3x + y^2 + 2y = 0$

Step 3: Substitute the given point into the circumcircle equation. We are given that the point $(\alpha, \frac{1}{2})$ lies on the circumcircle of triangle OPQ. This means the coordinates of the point must satisfy the equation of the circumcircle. Substituting $x = \alpha$ and $y = \frac{1}{2}$ into the equation $x^2 - 3x + y^2 + 2y = 0$: $\alpha^2 - 3\alpha + (\frac{1}{2})^2 + 2(\frac{1}{2}) = 0$ $\alpha^2 - 3\alpha + \frac{1}{4} + 1 = 0$ $\alpha^2 - 3\alpha + \frac{5}{4} = 0$

Step 4: Solve for $\alpha$. Multiply the equation by 4 to eliminate the fraction: $4\alpha^2 - 12\alpha + 5 = 0$ We can solve this quadratic equation by factoring: $(2\alpha - 1)(2\alpha - 5) = 0$ This gives us two possible solutions for $\alpha$: $2\alpha - 1 = 0 \Rightarrow \alpha = \frac{1}{2}$ $2\alpha - 5 = 0 \Rightarrow \alpha = \frac{5}{2}$

Step 5: Check against the options. The possible values for $\alpha$ are $\frac{1}{2}$ and $\frac{5}{2}$. The available options are: (A) 1 (B) $-\frac{1}{2}$ (C) $\frac{5}{2}$ (D) $\frac{3}{2}$ The value $\alpha = \frac{5}{2}$ matches option (C).

Common Mistakes & Tips

• Failing to recognize that OC is the diameter of the circumcircle makes the problem much more difficult.
• Errors in algebraic manipulation, especially when dealing with fractions and the quadratic formula.
• Forgetting the formula for the equation of a circle given the endpoints of a diameter.

Summary

We found the center of the given circle, determined the equation of the circumcircle of triangle OPQ, and then used the given point to solve for $\alpha$. The quadratic equation yielded two possible values for $\alpha$, and one of them matched one of the answer options.

The final answer is \boxed{\frac{5}{2}}, which corresponds to option (C).