AB is an external tangent to $S_1$ at A and to $S_2$ at B and common tangent at P cuts AB at Q.

Question

Question:

Two circles $S_1$ and $S_2$ of radius $3$ and $4$ touch each other externally at point P. If AB is an external tangent to $S_1$ at A and to $S_2$ at B and common tangent at P cuts AB at Q. Then, which of the following is/are correct?

A) $\frac{AP}{BQ}=\frac6{\sqrt{21}}$

B) $\frac{AP}{BP}=\frac{√3}{\sqrt{9}}$

C) Q is circumcentre of $\triangle PAB$

D) Area of $\triangle PAB=\frac{96√3}{7}$

Solution:

AQ=PQ=QB

$\implies$ Q is circumcentre of $\triangle PAB$

$\implies \frac{AQ}{QB}=\frac{PC_1}{PC_2}=\frac34$

$\cos2\theta=\frac{3^2+3^2-AP^2}{2(3)(3)}$

$\cos(π-2\theta)=\frac{4^2+4^2-BP^2}{2(4)(4)}$

$\implies 1-\frac{AP^2}{18}+1-\frac{BP^2}{32}=0$

Now, $AB=2\sqrt{3(4)}=4√3$

$\implies AQ=BQ=PQ=2√3$

And $AP^2+BP^2=AB^2\implies AP^2+BP^2=48$

$\implies AP=\frac{12}{√7}, BP=8\sqrt{\frac37}$

My Doubts:

1. How is $\angle AC_1P=2\theta?$

2. There seems to be some typo in $\frac{AQ}{QB}=\frac{PC_1}{PC_2}=\frac34$. How to correct it?