Let the tangents at two points A and B on the circle x 2 + y 2 - 4 x + 3 = 0 meet at origin O 0 , 0 . Then…

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 $x\times 0+y\times 0-4\left(\frac{x+0}{2}\right)+3=0$

$\Rightarrow AB:2x=3$

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

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

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

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