lf the tangents $\mathrm{P}\mathrm{A}$ and $\mathrm{P}\mathrm{B}$ are drawn from the point $\mathrm{P}(-1,2)$ to the circles $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}-2\mathrm{y}-3=0$ and $\mathrm{C}$ is the centre of the circle, then the area of the quadrilateral P$\mathrm{A}\mathrm{C}\mathrm{B}$ is

The centre of the circle C is (-1/2, 1) and its radius is sqrt(1/4 + 1 + 3) = sqrt(17/4). The distance from point P(-1,2) to the centre C is sqrt(1/4 + 1) = sqrt(5/4). Since the distance from P to the centre is less than the radius of the circle, point P lies inside the circle. Tangents cannot be drawn from an internal point to a circle, so the quadrilateral PACB and its area do not exist.

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