JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Circle

Rewrite the circle equation in standard form: $(x-2{)}^2+y^2=1$, so the center is $C(2,0)$ and radius $r=1$. Since tangents from the origin meet the circle at A and B, the chord of contact AB has equation $2x=3$ (using the formula $T=0$ from point $(0,0)$). The perpendicular distance from O to the line $x=\frac{3}{2}$ is $\mathrm{OM}=\frac{3}{2}$. Using the right triangle $\mathrm{OCA}$, we have $\mathrm{OA}=\sqrt{{\mathrm{OC}}^2-r^2}=\sqrt{4-1}=\sqrt{3}$. The half-length of chord AB is $\mathrm{AM}=\sqrt{{\mathrm{OA}}^2-{\mathrm{OM}}^2}=\sqrt{3-\frac{9}{4}}=\frac{\sqrt{3}}{2}$. Therefore, the area of triangle OAB is $\frac{1}{2}\times 2\mathrm{AM}\times \mathrm{OM}=\frac{1}{2}\times \sqrt{3}\times \frac{3}{2}=\frac{3\sqrt{3}}{4}$ sq. units.