Where do I look at this from? (2)

Geometry Level 5

Consider a circle \(S_{1}\),

\[x^{2}+y^{2}+1-2x(\cos\theta - \sin\theta)-2y(\cos\theta + \sin\theta)=0\]

Now, consider another circle \(S_{2}\), centred at \(P=(-\cos\theta - \sin\theta, \cos\theta - \sin\theta) \) such that \(S_{1}\) internally touches \(S_{2}\).

Draw a pair of tangents \(T_{1} \) and \(T_{2}\) from \(P\) to \(S_{1}\). Let these tangents meet the circle \(S_{2}\) at points \(A\) and \(B\) as shown. From a point \(R\) on \(S_{2}\), draw chords \(RA\) and \(RB\) with lengths \(l_{1} , l_{2}\), respectively to \(S_{2}\).

Then, there exists a certain \(\alpha\) such that one of

\[\large \left| \cos^{-1}\frac{l_{1}}{6} + \cos^{-1}\frac{l_{2}}{6}\right| \quad \text{ or } \quad \left|\cos^{-1}\frac{l_{1}}{6}-\cos^{-1}\frac{l_{2}}{6} \right|\]

is equal to \(\alpha\) regardless of the value of \(R\). Find the value of this constant \(\alpha\).


If you are looking for more such simple but twisted questions, twisted problems for jee aspirants is for you!
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