Question

The angle between the tangents from $\alpha ,\beta$ to the circle $x^2+y^2=a^2$ is

• A. ${\tan }^{-1}\left(\frac{a}{\sqrt{S_1}}\right)$
• B. $2{\tan }^{-1}\left(\frac{a}{\sqrt{S_1}}\right)$
• C. $2{\tan }^{-1}\left(\frac{\sqrt{S_1}}{a}\right)$
• D. $Noneofthese$

Answer 1

Answer: (C)

$2{\tan }^{-1}\left(\frac{\sqrt{S_1}}{a}\right)$

Given, radius of circle =a and center (0,0)

$C{O}^2=A{O}^2+A{C}^2$

$\left(\sqrt{(\propto -0{)}^2+(\beta -0{)}^2{)}^2}\right)=a^2+A{C}^2$

${\propto }^2+{\beta }^2-a^2=A{C}^2\Rightarrow AC=\sqrt{{\propto }^2+{\beta }^2-a^2}=\sqrt{5_1}$

In $△COA$

$\Rightarrow \tan \frac{\theta }{2}=\frac{AO}{AC}=\frac{a}{\sqrt{5_1}}$

$\frac{\theta }{2}={\tan }^{-1}\left[\frac{a}{\sqrt{5_1}}\right]$

$\theta =2{\tan }^{-1}\left[\frac{a}{\sqrt{5_1}}\right]$