Let the tangents at two points A and B on the circle $$x^2 + y^2 - 4x + 3 = 0$$ meet at origin $$O(0,0)$$. Then the area of the triangle OAB is

We have the circle $$x^2 + y^2 - 4x + 3 = 0$$. Rewriting in standard form: $$(x-2)^2 + y^2 = 1$$. So the centre is $$C(2, 0)$$ and radius $$r = 1$$.

The tangents from the origin $$O(0,0)$$ touch the circle at points A and B. The distance from the origin to the centre is $$OC = 2$$.

The length of the tangent from O to the circle is $$OA = OB = \sqrt{OC^2 - r^2} = \sqrt{4 - 1} = \sqrt{3}$$.

Now, in triangle OAC, we have $$OC = 2$$, $$CA = r = 1$$, and $$OA = \sqrt{3}$$. The angle $$\angle AOC$$ can be found using $$\sin(\angle AOC) = \frac{CA}{OC}$$... Actually, since OA is tangent to the circle at A, the angle $$\angle OAC = 90°$$.

So $$\sin(\angle AOC) = \frac{CA}{OC} = \frac{1}{2}$$, giving $$\angle AOC = 30°$$.

By symmetry, $$\angle BOC = 30°$$ as well, so $$\angle AOB = 60°$$.

The area of triangle OAB can be computed as:

Hence, the correct answer is Option 2.