$\mathrm{A}\mathrm{B}$ is a chord of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=9$. The tangent at $\mathrm{A}$ and $\mathrm{B}$ intersect at $\mathrm{C}$. lf $(1,2)$ is the midpoint of $\mathrm{A}\mathrm{B}$, then the area of $\Delta \mathrm{A}\mathrm{B}\mathrm{C}$ is (in square units)

Using the property of the chord with a given midpoint (h, k), the equation of chord AB is x*1 + y*2 - 9 = 1^2 + 2^2 - 9, which simplifies to x + 2y = 5. The perpendicular distance from the circle's centre (0,0) to the line x + 2y = 5 is d = 5 / sqrt(5) = sqrt(5). The radius of the circle is r = 3, so half the length of the chord AB is sqrt(r^2 - d^2) = sqrt(9 - 5) = 2, making the base AB equal to 4 units. The height of triangle ABC from vertex C to the line AB is equal to the distance from the chord to the tangent, which is calculated as 4/sqrt(5). Therefore, the area of triangle ABC is 1/2 * 4 * 4/sqrt(5) = 8/sqrt(5) square units.

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