Circle And System Of Circles Question 233
Question: From any point on the circle $ x^{2}+y^{2}=a^{2} $ tangents are drawn to the circle $ x^{2}+y^{2}=a^{2}{{\sin }^{2}}\alpha $ , the angle between them is
[RPET 2002]
Options:
A) $ \frac{\alpha }{2} $
B) $ \alpha $
C) $ 2\alpha $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let any point on the circle $ x^{2}+y^{2}=a^{2} $ be $ (a\cos t,a\sin t) $ and $ \angle ,OPQ=\theta $
Now; $ PQ= $ length of tangent from P on the circle $ x^{2}+y^{2}=a^{2}{{\sin }^{2}}\alpha $
$ \therefore $ $ PQ= $
$ \sqrt{a^{2}{{\cos }^{2}}t+a^{2}{{\sin }^{2}}t-a^{2}{{\sin }^{2}}\alpha } $
$ =a\cos \alpha $
$ OQ= $ Radius of the circle $ x^{2}+y^{2}=a^{2}{{\sin }^{2}}\alpha $
$ OQ= $
$ a\sin \alpha $ ,
$ \therefore $ $ \tan \theta =\frac{OQ}{PQ}=\tan \alpha \Rightarrow ,\theta =\alpha $
$ \therefore $ Angle between tangents $ =,\angle ,QPR=2\alpha . $
BETA
