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 $BC$ subtends an angle on the major arc $BC$ 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 $4x+3=0\Rightarrow x=-3/4$

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 $\tan \theta =\frac{BL}{C_{3}L}=\frac{\frac{3\sqrt{7}}{4}}{(3-3/4)}=\frac{\sqrt{7}}{3}$
$\Rightarrow \cos \theta =\frac{3}{4}$