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Two tangents are drawn from the point $P(-1,1)$ to the circle $x^{2}+y^{2}-2 x-6 y+6=0.$ If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $\mathrm{AB}$ and $\mathrm{AD}$ are equal, then the area of the triangle ABD is equal to
Option: 1 $2$
Option: 2 $\left ( 3\sqrt{2}+2 \right )$
Option: 3 $4$
Option: 4 $3\left ( \sqrt{2}-1 \right )$

Answers (1)

best_answer

$x^{2}+y^{2}-2 x-6 y+6=0$

$centre (1,3), radius =\sqrt{1+9-6}=2$

Clearly, $A$ will be $(1,1) \& B(-1,3)$ $\& P B \perp P A ., D(3,3)$
Area of $\triangle A B D=\frac{1}{2} \cdot C A \cdot B D$
$=\frac{1}{2} \cdot 2 \cdot 4=4$

Hence option (3) is correct.

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Kuldeep Maurya

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