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Question

From any point P on the circle $$\displaystyle x^{2}+y^{2}=9$$ two tangents are drawn to circle $$\displaystyle x^{2}+y^{2}=4$$ which cut $$\displaystyle x^{2}+y^{2}=9$$ at A and B then locus of point of intersection of tangents drawn $$\displaystyle x^{2}+y^{2}=9$$ at A and B is the curve


A
x2+y2=27
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B
x2+y2=272
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C
x2+2y2=272
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D
x2+y2=8
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Solution

The correct option is C $$\displaystyle x^{2}+y^{2}=27^{2}$$
We have $$ \displaystyle \sin \theta =\frac{2}{3}\Rightarrow \tan \theta =\frac{2}{\sqrt{5}} $$ 
$$ \displaystyle \cos 2\theta =\frac{3}{OR} $$

$$ \displaystyle \Rightarrow \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta }=\frac{3}{OR}\Rightarrow \frac{1-\dfrac{4}{5}}{1+\dfrac{4}{5}}=\frac{3}{OR} $$
$$ \displaystyle \Rightarrow OR=27 $$

$$ \displaystyle \Rightarrow  $$R is situated at a fixed distance from O 

$$ \displaystyle \Rightarrow  $$ Locus $$ \displaystyle x^{2}+y^{2}=27^{2} $$

Maths

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