Let the tangents at two points A and B on the circle x ^ 2 + y ^ 2...

Given the tangents at two points $A$ and $B$ on the circle ${x}^{2} + {y}^{2} - 4x + 3 = 0$ meet at origin $O\left(0, 0\right)$,

Plotting the diagram we get,

Now circle $C: {\left(x - 2\right)}^{2} + {y}^{2} = 1$

Equation of chord $AB$ is given by $T = 0$ i.e $x \times 0 + y \times 0 - 4\left(\dfrac{x + 0}{2}\right) + 3 = 0$

$⇒ AB: 2x = 3$

Now $OA = OB = \sqrt{{0}^{2} + {0}^{2} - 4 \times 0 + 3} = \sqrt{3}$

So, $AM = \sqrt{O{A}^{2} - O{M}^{2}} = \sqrt{3 - \dfrac{9}{4}} = \sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}$

Now area of triangle $OAB = \dfrac{1}{2}\left(2AM\right)\left(OM\right)$

$= \dfrac{3\sqrt{3}}{4}$ sq. units