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
B
x2+y2=272
C
x2+2y2=272
D
x2+y2=8

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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