Two tangents are drawn from the point P(- 1, 1) to the circle x ^ 2...

Given, the equation of circle is ${x}^{2} + {y}^{2} - 2x - 6y + 6 = 0$

So, the centre of the circle is \left(1, 3\right) and the radius is $\sqrt{{1}^{2} + {3}^{2} - 6} = 2$

Assume, the coordinate of point $B$ is \left(- 1, k\right)$and the coordinate of point A$ is \left(h, 1\right)$.</p><p>So,$\sqrt{{\left(h + 1\right)}^{2}} = 2 ⇒ h = 1$and$\sqrt{{\left(k - 1\right)}^{2}} = 2 ⇒ k = 3$</p><p>So, the coordinate of$B$is \left(- 1, 3\right)$ and the coordinate of A$is \left(1, 1\right) .</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/3eae8af9-e3a9-4920-a1d7-c0b84d91a29b-image.png" style="width: 284px; height: 222px;" /></p><p>From$ the diagram, area of∆ ABD = $\dfrac{1}{2}$ \times 2 \times 4$</p><p>$= 4$sq. units</p>$