Question

Two circles are
$S_1 \equiv(x+3)^2+y^2=9 $
$ S_2 \equiv(x-5)^2+y^2=16$
with centres $C_1 \& C_2$
Locus of circle cuts the circle $S_1$ at $B \& C$, then line segment $B C$ subtends an angle on the major arc $B C$ of circle $S_1$ which is

• A. $\cos ^{-1} \frac{3}{4}$
• B. $\tan ^{-1}(3 \sqrt{7})$
• C. $\cos ^{-1}\left(\frac{4}{3}\right)$
• D. $\cot ^{-1}\left(\frac{\sqrt{7}}{3}\right)$

Answer 1

Answer: (A)

Common chord of $S_1=0$ and $S_3=0$ is $4 x+3=0 \Rightarrow x=-3 / 4$

At $ x=-3 / 4, \left(-\frac{3}{4}+3\right)^2+y^2=9 $
$ \Rightarrow y^2=9-\frac{81}{16}$
$ \Rightarrow y^2=\frac{63}{16}$
$ \Rightarrow y=\pm \frac{3 \sqrt{7}}{4}$
$ \Rightarrow B \equiv\left(\frac{-3}{4}, \frac{3 \sqrt{7}}{4}\right) $
and $ C \equiv\left(\frac{-3}{4}, \frac{-3 \sqrt{7}}{4}\right)$
Hence $\tan \theta=\frac{B L}{C_3 L}=\frac{\frac{3 \sqrt{7}}{4}}{(3-3 / 4)}=\frac{\sqrt{7}}{3}$
$ \Rightarrow \cos \theta=\frac{3}{4}$