Two tangents are drawn from point mathrm{P}(-1,1) to circle mathrm{x}^{2}+mathrm{y}^{2}-2 mathrm{x}-

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10-1.Circle and System of Circles

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Two tangents are drawn from the point $\mathrm{P}(-1,1)$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-6 \mathrm{y}+6=0$. 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 $A B$ and $A D$ are equal, then the area of the triangle $A B D$ is eqaul to:

A

$2$

B

$(3 \sqrt{2}+2)$

C

$4$

D

$3(\sqrt{2}-1)$

(JEE MAIN-2021)

Solution

$\triangle \mathrm{ABD}=\frac{1}{2} \times 2 \times 4$

$=4$

Standard 11

Mathematics

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