Let $$O$$ be the origin and $$OP$$ and $$OQ$$ be the tangents to the circle $$x^2 + y^2 - 6x + 4y + 8 = 0$$ at the points $$P$$ and $$Q$$ on it. If the circumcircle of the triangle $$OPQ$$ passes through the point $$\left(\alpha, \frac{1}{2}\right)$$, then a value of $$\alpha$$ is

A key property in geometry is that the circumcircle of the triangle formed by an external point and its points of tangency ($$OPQ$$) always has the line segment joining the external point ($$O$$) and the center of the circle ($$C$$) as its diameter.

1. Find the Circle's Center

The given circle is $$x^2 + y^2 - 6x + 4y + 8 = 0$$.

By comparing with $$x^2 + y^2 + 2gx + 2fy + c = 0$$, we get:

• Center $$C = (-g, -f) = (3, -2)$$

2. Equation of the Circumcircle of $$\triangle OPQ$$

Point $$O$$ is $$(0,0)$$ and Point $$C$$ is $$(3,-2)$$. Since $$OC$$ is the diameter, we use the diameter form:

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

$$(x - 0)(x - 3) + (y - 0)(y + 2) = 0$$

$$x^2 - 3x + y^2 + 2y = 0$$

3. Solve for $$\alpha$$

The circle passes through $$(\alpha, \frac{1}{2})$$. Substitute these coordinates:

$$\alpha^2 - 3\alpha + \left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) = 0$$

$$\alpha^2 - 3\alpha + \frac{1}{4} + 1 = 0$$

$$\alpha^2 - 3\alpha + \frac{5}{4} = 0$$

Multiply by 4:

$$4\alpha^2 - 12\alpha + 5 = 0$$

Factor the quadratic:

$$(2\alpha - 5)(2\alpha - 1) = 0$$

$$\alpha = \frac{5}{2} \text{ or } \alpha = \frac{1}{2}$$

Comparing with the options, the value of $$\alpha$$ is $$\frac{5}{2}$$.

Correct Option: (C)