If $\theta$ is the angle between the tangents from $(-1,0)$ to the circle $x^2+y^2-5 x+4 y-2=0$, then…
If $\theta$ is the angle between the tangents from $(-1,0)$ to the circle $x^2+y^2-5 x+4 y-2=0$, then $\theta$ is equal to
$2 \tan ^{-1}\left(\frac{7}{4}\right)$
$\tan ^{-1}\left(\frac{7}{4}\right)$
$2 \cot ^{-1}\left(\frac{7}{4}\right)$
$\cot ^{-1}\left(\frac{7}{4}\right)$
Solution
We know that, the angle between the two tangents from $(\alpha, \beta)$ to the circle $x^2+y^2=r^2$ is
$
2 \tan ^{-1} \frac{r}{\sqrt{S_1}}
$
Given equation of circle is
$
x^2+y^2-5 x+4 y-2=0 \text {. }
$
Now, radius, $r=\sqrt{\left(-\frac{5}{2}\right)^2+(2)^2+2}=\sqrt{\frac{49}{4}}$
$
=\frac{7}{2}
$
At point $(-1,0)$
$
\begin{aligned}
S_1 & =(-1)^2+(0)^2-5(-1)+4(0)-2 \\
& =1+5-2=4
\end{aligned}
$
$\therefore$ Required angle, $\theta=2 \tan ^{-1} \frac{\frac{7}{2}}{\sqrt{4}}$
$
=2 \tan ^{-1}\left(\frac{7}{4}\right)
$