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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
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Verified Answer
The correct answer is:
$2 \tan ^{-1}\left(\frac{7}{4}\right)$
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, $\theta=2 \tan ^{-1} \frac{\frac{7}{2}}{\sqrt{4}}$
$$
=2 \tan ^{-1}\left(\frac{7}{4}\right)
$$
$$
2 \tan ^{-1} \frac{r}{\sqrt{S_1}}
$$
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, $\theta=2 \tan ^{-1} \frac{\frac{7}{2}}{\sqrt{4}}$
$$
=2 \tan ^{-1}\left(\frac{7}{4}\right)
$$
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