If the tangents drawn at the points \( O(0,0) \) and \( P(1+\sqrt{5}, 2) \) on the circle \( x^2 + y^2 - 2x - 4y = 0 \) intersect at the point \( Q \), then the area of the triangle \( OPQ \) is equal to:
\( \frac{3 + \sqrt{5}}{2} \)
\( \frac{4 + 2\sqrt{5}}{2} \)
\( \frac{5 + 3\sqrt{5}}{2} \)
\( \frac{7 + 3\sqrt{5}}{2} \)