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Let the tangent to the circle \mathrm{C}_{1}: x^{2}+y^{2}=2 at the point \mathrm{M}(-1,1) intersect the circle \mathrm{C}_{2}:(x-3)^{2}+(y-2)^{2}=5, at two distinct points \mathrm{A}$ and $\mathrm{B}. If the tangents to \mathrm{C}_{2} at the points \mathrm{A}$ and $\mathrm{B} intersect at N, then the area of the triangle ANB is equal to :

Option: 1

\frac{1}{2}


Option: 2

\frac{2}{3}


Option: 3

\frac{1}{6}


Option: 4

\frac{5}{6}


Answers (1)

best_answer

\mathrm{Using \: T=0, equation \: of\: tangent \: at \: M(-1,1)\: to \: x^{2}+y^{2}=2 \: is}

\mathrm{x x_{1}+y y_{1}=2} \\

\mathrm{-x+y=2} \\

\mathrm{\Rightarrow y=x+2}

 

\mathrm{\text { CM }=\frac{|2-3-2|}{\sqrt{1+1}}=\frac{3}{\sqrt{2}}}\\

\mathrm{Now \cos \theta=\frac{C M}{B C}=\frac{3}{\sqrt{2} \cdot \sqrt{5}}=\frac{3}{\sqrt{10}}}

\mathrm{\text { And } \quad B M =\sqrt{C B^{2}-C M^{2}}} \\

\mathrm{=\sqrt{5-\frac{9}{2}} }\\

\mathrm{=\frac{1}{\sqrt{2}}}

\mathrm{Area \: of \: \triangle A N B=2 \: area\: of\: \triangle N M B }

\mathrm{=2 \times \frac{1}{2} \times M B \cdot M N }\\

\mathrm{=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{M B}{\tan (90-\theta)}\right) }\\

\mathrm{=\frac{1}{2} \cdot \tan \theta} \\

\mathrm{=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}}

Hence the correct answer is option 3.

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vinayak

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