Circle And System Of Circles Question 12
Question: The angle between the tangents from $ (\alpha ,\beta ) $ to the circle $ x^{2}+y^{2}=a^{2} $ , is
Options:
A) $ {{\tan }^{-1}}( \frac{a}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}}} ) $
B) $ {{\tan }^{-1}}( \frac{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}}}{a} ) $
C) $ 2{{\tan }^{-1}}( \frac{a}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}}} ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tan \frac{\theta }{2}=\frac{CT_1}{PT_1}=\frac{a}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}}} $
$ \frac{\theta }{2}={{\tan }^{-1}}\frac{a}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}}}\Rightarrow \theta =2{{\tan }^{-1}}\frac{a}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}-a^{2}}} $ .
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