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(A) The angle $	heta$ between the two tangents drawn from an external point $P(x_1, y_1)$ to a circle with radius $r$ is given by $	heta = 2 	an^{-

Vedclass · Accepted Answer

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

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(A) The angle $	heta$ between the two tangents drawn from an external point $P(x_1, y_1)$ to a circle with radius $r$ is given by $	heta = 2 	an^{-1}\left(\frac{r}{\sqrt{S_1}}ight)$,where $S_1$ is the power of the point with respect to the circle.Given the circle equation $x^2+y^2-5x+4y-2=0$,we compare it with $x^2+y^2+2gx+2fy+c=0$ to find $g = -\frac{5}{2}$,$f = 2$,and $c = -2$.The radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{\left(-\frac{5}{2}ight)^2 + 2^2 - (-2)} = \sqrt{\frac{25}{4} + 4 + 2} = \sqrt{\frac{25+24}{4}} = \sqrt{\frac{49}{4}} = \frac{7}{2}$.For the point $P(-1, 0)$,the power of the point $S_1$ is $(-1)^2 + (0)^2 - 5(-1) + 4(0) - 2 = 1 + 5 - 2 = 4$.Thus,$	heta = 2 	an^{-1}\left(\frac{7/2}{\sqrt{4}}ight) = 2 	an^{-1}\left(\frac{7/2}{2}ight) = 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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