The angle between the two tangents drawn from | vedclass

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(B) Let $PT$ and $PQ$ be the tangents drawn from point $P(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$,and let $\angle TPQ = 	heta$.If $O$ i

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$2	an^{-1}\left(\frac{a}{\sqrt{S_1}}ight)$

Vedclass · Answer

(B) Let $PT$ and $PQ$ be the tangents drawn from point $P(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$,and let $\angle TPQ = 	heta$.If $O$ is the center of the circle,then $\angle TPO = \angle QPO = 	heta/2$.In the right-angled triangle $	riangle OTP$,we have:$	an(	heta/2) = \frac{OT}{PT} = \frac{a}{\sqrt{S_1}}$,where $S_1 = \alpha^{2} + \beta^{2} - a^{2}$.Therefore,$	heta/2 = 	an^{-1}\left(\frac{a}{\sqrt{S_1}}ight)$.Hence,$	heta = 2	an^{-1}\left(\frac{a}{\sqrt{S_1}}ight)$.

The angle between the two tangents drawn from the point $(\alpha, \beta)$ to the circle $x^{2} + y^{2} = a^{2}$ is:

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