Two tangents are drawn from the point P(-1, 1 | vedclass

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(C) The equation of the circle is $x^{2}+y^{2}-2x-6y+6=0$,which can be written as $(x-1)^{2}+(y-3)^{2}=2^{2}$. The center is $C(1, 3)$ and the radius

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$4$

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(C) The equation of the circle is $x^{2}+y^{2}-2x-6y+6=0$,which can be written as $(x-1)^{2}+(y-3)^{2}=2^{2}$. The center is $C(1, 3)$ and the radius $r=2$.From the point $P(-1, 1)$,the tangents touch the circle at $A(1, 1)$ and $B(-1, 3)$.The length of the chord $AB$ is $\sqrt{(1 - (-1))^{2} + (1 - 3)^{2}} = \sqrt{2^{2} + (-2)^{2}} = \sqrt{8} = 2\sqrt{2}$.Given that the length of segment $AD$ is equal to $AB$,$AD = 2\sqrt{2}$.In the circle,the length of a chord is given by $2\sqrt{r^{2}-d^{2}}$,where $d$ is the distance from the center to the chord. For $AD = 2\sqrt{2}$,$2\sqrt{2^{2}-d^{2}} = 2\sqrt{2} \implies 4-d^{2}=2 \implies d^{2}=2 \implies d=\sqrt{2}$.The distance from center $C(1, 3)$ to chord $AD$ is $\sqrt{2}$. The distance from $C$ to $AB$ is also $\sqrt{2}$.Since $AB$ and $AD$ are chords of equal length,they are equidistant from the center. The area of $	riangle ABD$ with base $AB = 2\sqrt{2}$ and height $h$ (distance from $D$ to $AB$) is calculated as $4$ square units.

Two tangents are drawn from the point $P(-1, 1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$. If these tangents touch the circle at points $A$ and $B$,and if $D$ is a point on the circle such that the lengths of the segments $AB$ and $AD$ are equal,then the area of the triangle $ABD$ is equal to:

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