The angle between the tangents drawn from the origin to the circle $x^2+y^2+4 x-6 y+4=0$ is
- $\tan ^{-1}\left(\frac{5}{13}\right)$
- $\tan ^{-1}\left(\frac{5}{12}\right)$
- $\tan ^{-1}\left(\frac{-12}{2}\right)$
- $\tan ^{-1}\left(\frac{13}{2}\right)$
Solution
$\text { Centre } \equiv(-g,-f) \equiv(-2,3)$ Radius of circle, $\mathrm{OA}=\sqrt{g^2+f^2-c}$ $\begin{aligned} & =\sqrt{(2)^2+(-3)^2-4} \\ & =\sqrt{4+9-4}=3 \end{aligned}$ Length of tangent $\begin{aligned} & \mathrm{PA}=\mathrm{PB} \\ & =\sqrt{(0)^2+(0)^2+4(0)-6(0)+4} \\ & =\sqrt{4}=2 \end{aligned}$ In $\triangle \mathrm{OAP}, \tan \theta=\frac{O A}{P A}=\frac{3}{2}$ Now, the angle between the tangent drawn from origin $\begin{aligned} & 2 \theta=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right) \\ & 2 \theta=\tan ^{-1}\left(\frac{2 \times \frac{3}{2}}{1-\frac{9}{4}}\right) \\ & \Rightarrow \quad 2 \theta=\tan ^{-1}\left(\frac{-12}{5}\right) \end{aligned}$
Asked in: AP EAMCET 2016