Question

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

• A. $ta{n}^{-1}\frac{a}{\sqrt{{\alpha }^2+{\beta }^2-a^2}}$
• B. $ta{n}^{-1}\frac{\sqrt{{\alpha }^2+{\beta }^2-a^2}}{a}$
• C. $2ta{n}^{-1}\frac{a}{\sqrt{{\alpha }^2+{\beta }^2-a^2}}$
• D. $Noneofthese$

Answer 1

The correct option is C $2ta{n}^{-1}\frac{a}{\sqrt{{\alpha }^2+{\beta }^2-a^2}}$

$tan=\frac{\theta }{2}=\frac{C{T}_1}{P{T}_1}=\frac{a}{\sqrt{{\alpha }^2+{\beta }^2-a^2}}$

$\frac{\theta }{2}=ta{n}^{-1}\frac{a}{\sqrt{{\alpha }^2+{\beta }^2-a^2}}\Rightarrow \theta =2ta{n}^{-1}\frac{a}{\sqrt{{\alpha }^2+{\beta }^2-a^2}}$.