Circle And System Of Circles Question 257
Question: The line $ (x-a)\cos \alpha +(y-b) $ $ \sin \alpha =r $ will be a tangent to the circle $ {{(x-a)}^{2}}+{{(y-b)}^{2}}=r^{2} $
Options:
A) If $ \alpha =30^{o} $
B) If $ \alpha =60^{o} $
C) For all values of $ \alpha $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
According to the condition of tangency
$ r=\frac{a\cos \alpha +b\sin \alpha -(a\cos \alpha +b\sin \alpha )-r}{\sqrt{{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha }} $
$ \Rightarrow r=|-r|\ \Rightarrow r=r $ . Therefore, it is a tangent to the circle for all values ofa.
BETA
