If $ heta$is the angle between the tangents from $( - 1,\ 0)$ to the circle $x^{2} + y^{2} - 5x + 4y - 2 = 0$, then $ heta$ is equal to

Question

If $	heta$is the angle between the tangents from $( - 1,\ 0)$ to the circle $x^{2} + y^{2} - 5x + 4y - 2 = 0$, then $	heta$ is equal to

Dr. Abubacker · Accepted Answer

(a)
We know, that, the angle between the two tangents from $(\alpha,\ \beta)$
To the circle $x^{2} + y^{2} = r^{2}$ is $2	an^{- 1}\frac{r}{\sqrt{S_{1}}}$
Let $S = x^{2} + y^{2} - 5x + 4y - 2$
Here, $r = \sqrt{\left( - \frac{5}{2} ight)^{2} + (2)^{2} + 2} = \frac{7}{2}$
At point $( - 1,\ 0)$
$S_{1} = ( - 1)^{2} + (0)^{2} - 5( - 1) + 4(0) - 2 = 4$
$	herefore$ Required angle, $	heta = 2	an^{- 1}\frac{7/2}{\sqrt{4}}$
$= 2	an^{- 1}\left( \frac{7}{4} ight)$

Q. If $\theta$is the angle between the tangents from $( - 1,\ 0)$ to the circle $x^{2} + y^{2} - 5x + 4y - 2 = 0$, then $\theta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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