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Question

(A) The angle $	heta$ between the two tangents from a point $P(x_1, y_1)$ to a circle $S=0$ is given by $	an\left(\frac{	heta}{2}ight) = \frac{r}

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$2 	an^{-1}\left(\frac{7}{4}ight)$

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(A) The angle $	heta$ between the two tangents from a point $P(x_1, y_1)$ to a circle $S=0$ is given by $	an\left(\frac{	heta}{2}ight) = \frac{r}{\sqrt{S_1}}$,where $r$ is the radius and $S_1$ is the power of the point.Given the circle equation $x^2+y^2-5x+4y-2=0$,the center is $(\frac{5}{2}, -2)$ and the radius $r = \sqrt{(\frac{5}{2})^2 + (-2)^2 - (-2)} = \sqrt{\frac{25}{4} + 4 + 2} = \sqrt{\frac{49}{4}} = \frac{7}{2}$.The power of the point $(-1, 0)$ is $S_1 = (-1)^2 + (0)^2 - 5(-1) + 4(0) - 2 = 1 + 5 - 2 = 4$.Thus,$	an\left(\frac{	heta}{2}ight) = \frac{7/2}{\sqrt{4}} = \frac{7/2}{2} = \frac{7}{4}$.Therefore,$\frac{	heta}{2} = 	an^{-1}\left(\frac{7}{4}ight)$,which implies $	heta = 2 	an^{-1}\left(\frac{7}{4}ight)$.

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

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