Circle And System Of Circles Question 38
Question: The circles $ x^{2}+y^{2}=9 $ and $ x^{2}+y^{2}-12y+27=0 $ touch each other. The equation of their common tangent is
[MP PET 1998]
Options:
A) $ \frac{2b}{\sqrt{a^{2}-4b^{2}}} $
B) $ \frac{\sqrt{a^{2}-4b^{2}}}{2b} $
C) $ \frac{2b}{a-2b} $
D) $ \frac{b}{a-2b} $
Show Answer
Answer:
Correct Answer: A
Solution:
Any tangent to $ x^{2}+y^{2}=b^{2} $ is $ y=mx-b,\sqrt{1+m^{2}}. $ It touches $ {{(x-a)}^{2}}+y^{2}=b^{2} $ , if $ \frac{ma-b\sqrt{1+m^{2}}}{\sqrt{m^{2}+1}}=b $ or $ ma=2b\sqrt{1+m^{2}} $ or $ m^{2}a^{2}=4b^{2}+4b^{2}m^{2} $ ,
$ \therefore $ $ m=\pm ,\frac{2b}{\sqrt{a^{2}-4b^{2}}} $ .
BETA
